Chapter 2 Homework Packet
Conceptual Questions
1) Suppose that an object travels from one point in space to another. Make a comparison between the displacement
and the distance traveled.
A) The displacement can be either greater than, smaller than, or equal to the distance traveled.
B) The displacement is always equal to the distance traveled.
C) The displacement is either greater than or equal to the distance traveled.
D) The displacement is either less than or equal to the distance traveled.
Remember that displacement is the difference between the final position and initial position,
both in magnitude and direction. However, the object does not necessarily travel between the
initial and final position using the shortest path. If it does, then the displacement and distance
traveled are the same. But, if the path deviates from a straight line between the initial and
final position, even just a little, the distance traveled will be greater than displacement. So, the
displacement is either less than or equal to the distance traveled. Answer D
2) When is the average velocity of an object equal to the instantaneous velocity?
A) only when the velocity is constant
B) only when the velocity is increasing at a constant rate
C) always
D) never
Δx x2 − x1
-­‐-­‐the total displacement =
Δt t2 − t1
divided by the total time interval. During this time interval, the actual velocity could have changed (or it might not have). In contrast, the instantaneous velocity is the velocity at any “instant” in time during the motion. The only time the instantaneous velocity will be the same as the average velocity is if the velocity is the same throughout the motion—that is, if it is a constant velocity. If this is the case, the instantaneous velocity at any point in the motion will be the same as whatever that constant velocity is. Answer A Remember that the average velocity is v =
3) A new car manufacturer advertises that their car can go "from zero to sixty in 8 s". This is a description of
A) instantaneous speed.
B) average acceleration.
C) average speed.
D) instantaneous acceleration.
Think—zero to 60 translates into v0 = 0 mph, v = 60 mph, and t = 8s so we have the values Δv
. A statement of Δt
instantaneous speed would be something like, “at the instant the car passed the officer, it was travelling at 60 miles per hour. A statement of average speed would require a distance and a time interval. A statement of this might be something like, “on their trip the family traveled 735 miles in ten and a half hours. A statement of instantaneous acceleration might be something like, “At the bottom of the carnival ride the passengers experienced an acceleration of 17.6 m/s2. Answer B that plug into the average acceleration equation, a =
1 4) An object moving in the +x axis experiences an acceleration of 2.0 m/s2. This means the object is
A) changing its velocity by 2.0 m/s.
B) increasing its velocity by 2.0 m/s in every second.
C) traveling at 2.0 m in every second.
D) traveling at 2.0 m/s in every second.
Remember that the definition of acceleration is change in velocity during a specific time
interval. So, it might be tempting to select A if you don’t read it carefully enough. The
statement, changing its velocity by 2.0 m/s would just simply mean a one time change in
velocity, for example, from 2.0 m/s to 4.0 m/s. An acceleration of 2.0 m/s2 means a change in
velocity (with the positive value this would mean an increase in velocity) of 2.0 m/s for every
second that the acceleration occurs. If the beginning velocity was 2 m/s, after 1 second of
acceleration the velocity would be 4.0 m/s, after 2 s, 6.0 m/s, etc. In answers C and D,
“traveling at 2.0 m in every second, and traveling at 2.0 m/s in every second basically say the
same thing—a constant velocity of 2.0 m/s—so, not an acceleration. Answer B
5) Suppose that a car traveling to the East (+x direction) begins to slow down as it approaches a traffic light. Make
a statement concerning its acceleration.
A) The car is decelerating, and its acceleration is negative.
B) The acceleration is zero.
C) The car is decelerating, and its acceleration is positive.
D) A statement cannot be made using the information given.
For questions asking about direction of acceleration and velocity, always have a picture of a motion diagram in your mind (or you could draw one if you need to) visualizing what the velocity vectors and the acceleration vector are doing. In this case, the motion is in the positive direction but velocity is decreasing (deceleration) and the lengths of the velocity vectors must be decreasing. Remember then our rule that the acceleration vector must be in the direction of increasing lengths of the velocity vectors, so, the acceleration must be in the negative direction to cause this. Answer A
6) Suppose that a car traveling to the West (-x direction) begins to slow down as it approaches a traffic light. Make
a statement concerning its acceleration.
A) The acceleration is zero.
B) A statement cannot be made using the information given.
C) The car is decelerating, and its acceleration is positive.
D) The car is decelerating, and its acceleration is negative.
For questions asking about direction of acceleration and velocity, always have a picture of a motion diagram in your mind (or you could draw one if you need to) visualizing what the velocity vectors and the acceleration vector are doing. In this case, the motion is in the negative direction, and, velocity is decreasing (deceleration), so the lengths of the velocity vectors must be decreasing. Remember then our rule that the acceleration vector must be in the direction of increasing lengths of the velocity vectors, so, the acceleration must be in the positive direction to cause this. Answer C 2 7) Suppose that an object is moving with a constant velocity. Make a statement concerning its acceleration.
A) The acceleration must be a constant non-zero value.
B) The acceleration must be equal to zero.
C) The acceleration must be constantly increasing.
D) The acceleration must be constantly decreasing.
Δv
. If the velocity is constant, the
t
change in velocity , Δv , must be 0. Therefore, the acceleration must also be 0. Answer B
Remember the mathematical definition of acceleration: a =
8) If the velocity of an object is zero, does it mean that the acceleration is zero? Support your answer with an
example.
A) no, and an example would be an object starting from rest
B) no, and an example would be an object coming to a stop
C) yes, because of the way in which velocity is defined
D) yes, because of the way in which acceleration is defined
Consider an object at rest. At that instant, its velocity is 0 m/s. If a force on the object then
causes an acceleration, at the instant that the acceleration begins, what ever the magnitude of
that acceleration, the object is under a non-zero acceleration, yet the velocity is 0 m/s. (You
might be tempted to make an argument for answer B. Consider a ball thrown upward in the
air. As the ball travels upward, the acceleration due to gravity is acting on the ball causing it
to decelerate and finally come to a stop (velocity of 0 m/s) at its maximum height. After an
instant, the ball then returns toward the ground at an increasing negative velocity. The entire
time of the flight, the acceleration due to gravity is acting on the ball at -9.8 m/s2. This
acceleration is constant—it is not zero and is not changing, yet at one point, the velocity was 0.
However, the problem states that the velocity is 0, not that the velocity becomes 0. Answer A
9) Can an object's velocity change direction when its acceleration is constant? Support your answer with an
example.
A) No, this is not possible because it is always speeding up.
B) Yes, this is possible, and a car that starts from rest, speeds up, slows to a stop, and then backs up is an example.
C) No, this is not possible because it is always speeding up or always slowing down, but it can never turn around.
D) Yes, this is possible, and a rock thrown straight up is an example.
Consider an object in motion in a positive direction under the influence of a constant negative
acceleration. The negative acceleration would initially cause the velocity of the object to
decrease and ultimately come to a stop at a rate consistent with the negative acceleration.
However, as the object would continue to be under the influence of the acceleration, it would
then reverse its direction of motion and move at an increasing rate in the opposite direction.
The answer in the above list that would most consistent with this is D, a rock thrown straight
up under the influence of the constant acceleration due to gravity. (B would not work as an
example because in that circumstance, the acceleration would be changing, not constant.)
Answer D
10) Suppose that an object is moving with constant acceleration. Make a statement concerning its motion with
respect to time.
A) In equal times its speed increases by equal amounts.
B) In equal times its velocity changes by equal amounts.
C) In equal times it moves equal distances.
D) A statement cannot be made using the information given.
3 Δv
. This definition specifically
t
states, “change in velocity per interval of time,” which is consistent with statement B.
Statement A would not be correct because it uses the term speed, not velocity, and also
specifies that the speed must be increasing. However, acceleration could involve increasing or
decreasing velocity. Statement C is not correct because it specifies distance, not displacement,
and specifies equal distances in equal times, which would be constant speed rather than
changing speed. As statement B is a correct statement, statement D cannot be a correct
statement. Answer B
Remember the mathematical definition of acceleration: a =
11) Can an object have increasing speed while its acceleration is decreasing? Support your answer with an example.
A) Yes, and an example would be an object falling in the absence of air friction.
B) No, this is impossible because of the way in which acceleration is defined.
C) No, because if acceleration is decreasing the object will be slowing down.
D) Yes, and an example would be an object released from rest in the presence of air friction.
This question is tricky because the correct answer could actually be worded better—but, there
are also other tricky parts. First, consider that saying that acceleration is decreasing means
that the acceleration is not constant. In an algebra based physics class we do not usually
consider situations of changing acceleration. Mathematically, considering changing
acceleration would require calculus—in an algebra based class we usually only consider
constant acceleration. However, even if we can’t mathematically talk about changing
acceleration, we can descriptively. To say that acceleration is decreasing might mean that an
object is under the influence of constant acceleration in one direction, but, that the affect of
this constant acceleration is being altered by additional forces.
Consider an object falling from a height. Although we usually ignore air friction in order to
make our problems more convenient, here we need to consider an object falling due to gravity,
but, that there is also an upward force due to air friction. The thing you probably don’t know
about air resistance, is that as an object falls at a faster and faster rate, the opposing force due
to air resistance increases. So, in the first part of the fall, air resistance is not as great, and
the acceleration due to gravity is not altered to as great a degree, and the velocity increases as
the object falls. However, as the velocity increases, the force of the air resistance increases
and while the velocity continues to increase, it increases at a slower and slower rate. So,
acceleration would be decreasing while velocity would still be increasing. But, the upward
force of resistance would eventually balance the downward force due to gravity, and there
would finally be no more acceleration. At that point the object would be travelling at a
constant velocity we call terminal velocity.
So, in the above list, A would be incorrect because what we just described would not happened
without air resistence—in that scenario, acceleration would remain constant, it would not
decrease. For B, the way we define acceleration does not really have any impact on whether
velocity can or cannot increase in the situation of decreasing acceleration. C is incorrect—we
have just described a situation in which the acceleration is decreasing, and yet the velocity is
still increasing—it is just increasing at a “decreasing rate.” D is correct—but it is poorly
worded. It is not obvious that the question is saying the object is released from rest so that it
“is falling” in the presence of air friction. Answer D
4 12) Suppose a can, after an initial kick, moves up along a smooth hill of ice. Make a statement concerning its
acceleration.
A) It will have a constant acceleration up the hill, but a different constant acceleration when it comes back down the
hill.
B) It will have the same acceleration, both up the hill and down the hill.
C) It will travel at constant velocity with zero acceleration.
D) It will have a varying acceleration along the hill.
When it is stated that an object moves along a smooth hill of ice, we are to assume that there is
basically no friction. This means that as the object moves up and down the hill of ice, the only
force influencing its velocity is the acceleration due to gravity. As this acceleration is a
constant acceleration—acceleration due to gravity does not change and is not being changed
by friction—the can will have the same acceleration both up and down the hill. While it is
traveling upward, this negative acceleration will cause its positive velocity to decrease and
eventually come to a 0 velocity, and then direction will reverse, and the negative acceleration
will cause the negative velocity of the can to increase. Answer B
13) Under what condition is average velocity equal to the average of the object's initial and final velocity?
A) The acceleration must be constantly changing.
B) The acceleration must be constant.
C) This can only occur if there is no acceleration.
D) This is impossible.
If velocity changes at a constant rate (which is the same thing as saying constant acceleration)
then the average velocity must equal the sum of the initial and final velocity divided by 2,
which is the average of the object’s initial and final velocities. Answer B
14) Objects A and B both start at rest. They both accelerate at the same rate. However, object A accelerates for
twice the time as object B. What is the final speed of object A compared to that of object B?
A) the same speed
B) twice as fast
C) three times as fast
D) four times as fast
When you have questions that ask you to compare how changing a variable will affect a
resulting variable, the best way to see this is by writing out the formula and then change the
affected variable in the manner describe and visualize what would happen to the resulting
variable. Here we are told an object starts at rest (v0 = 0 m/s), a is the same for both objects,
but t is two times greater for A than for B. We are asked, what happens to final velocity. So,
we have v0, a, and t, and we want v. The formula we would use to help us is
v = v0 + at. So, write out this formula, highlighting in some way that you want to know what
happens to v if you double time—I would write it out like this:
v = v0 + a 2 t → 2 v
I have indicated by drawing a box around the 2 in front of the t that the change we are making
is to double the time. Then I write after that, the affect this has on the velocity. This basically
shows, “if I double the time, I double the velocity.” You can waste large quantities of time
trying to reason something out like this in your head if you don’t have a visual cue in front of
you. I highly recommend this approach. Answer B
15) Objects A and B both start from rest. They both accelerate at the same rate. However, object A accelerates for
twice the time as object B. What is the distance traveled by object A compared to that of object B?
A) three times as far
B) four times as far
5 C) the same distance
D) twice as far
In this case we are given the same information as in 14, but asked to determine the effect on
displacement. In this case we have x0 = 0 m, v0 = 0 m/s, a, and we are given that time is
doubled, and we want x. The appropriate formula to help us here is x = x0 + v0t + ½ at2. Lets
see what happens to x when we double t. Write out this formula, highlighting in some way
that you want to know what happens to x if you double time—I would write it out like this:
x = x0 + v0t + 12 a 2 t 2 → 4 x
I have indicated by drawing a box around the 2 in front of the t that the change we are making
is to double the time. Then I write after that, the affect this has on the displacement. This
basically shows, “if I double the time, because the doubling is squared, I quadruple the
displacement.” You can waste large quantities of time trying to reason something out like this
in your head if you don’t have a visual cue in front of you. I highly recommend this approach.
Answer B
16) When an object is released from rest and falls in the absence of friction, which of the following is true
concerning its motion?
A) The speed of the falling object is proportional to its mass.
B) The speed of the falling object is proportional to its weight.
C) The speed of the falling object is inversely proportional to its surface area.
D) None of the above is true.
In the absence of any other forces, the only thing affecting the velocity of a falling object is the acceleration due to gravity. Neither mass nor weight have anything to do with this. Surface area will only come into play if there is an atmosphere, in which case the upward force on the surface area caused by the particles in the air will oppose acceleration due to gravity. None of the given answers is correct. Answer D
17) When an object is released from rest and falls in the absence of friction, which of the following is true
concerning its motion?
A) Its acceleration is constant.
B) Its velocity is constant.
C) Neither its acceleration nor its velocity is constant.
D) Both its acceleration and its velocity are constant.
See the explanation for question 17. Again, the only thing that will affect the motion of the
falling object in the absence of air resistance is the acceleration due to gravity. While the
velocity of the falling object will increase, the acceleration due to gravity is a constant—it does
not change. Answer A
18) Suppose a ball is thrown straight up. Make a statement about the velocity and the acceleration when the ball
reaches the highest point.
A) Both its velocity and its acceleration are zero.
B) Its velocity is zero and its acceleration is not zero.
C) Its velocity is not zero and its acceleration is zero.
D) Neither its velocity nor its acceleration is zero.
Remember that the acceleration due to gravity is -9.8 m/s2—its direction is downward. As the
ball is thrown upward, once it leaves the thrower’s hand the only thing influencing its motion
(unless otherwise stated—we usually ignore air resistance), is this acceleration due to gravity.
This will cause the ball to decelerate—that is, its velocity will decrease—until it comes to a
stop at its maximum height. However, while the velocity has changed, the acceleration has
6 not—it remains an acceleration of -9.8 m/s2. After the ball comes to a stop, the negative
acceleration will then cause the ball to reverse direction and increase in velocity in a negative
direction. As it reaches the point where it was thrown, the magnitude (value) of its negative
velocity will equal the magnitude of the positive velocity with which it was thrown. Again,
throughout the whole flight, while the velocity changes, the acceleration does not. So, the
correct answer is, at maximum height, the velocity is 0 and its acceleration is not zero.
Answer B
19) Suppose a ball is thrown straight up. What is its acceleration just before it reaches its highest point?
A) exactly g
B) zero
C) slightly less than g
D) slightly greater than g
Remember that the acceleration due to gravity is -9.8 m/s2—its direction is downward. As the
ball is thrown upward, once it leaves the thrower’s hand the only thing influencing its motion
(unless otherwise stated—we usually ignore air resistance), is this acceleration due to gravity.
This will cause the ball to decelerate—that is, its velocity will decrease—until it comes to a
stop at its maximum height. However, while the velocity has changed, the acceleration has
not—it remains an acceleration of -9.8 m/s2. After the ball comes to a stop, the negative
acceleration will then cause the ball to reverse direction and increase in velocity in a negative
direction. As it reaches the point where it was thrown, the magnitude (value) of its negative
velocity will equal the magnitude of the positive velocity with which it was thrown. Again,
throughout the whole flight, while the velocity changes, the acceleration does not. So, the
correct answer is, at a point just before it reaches its highest point, the acceleration is exactly
g—the same as it is at any point in its flight. Answer A
20) Suppose a ball is thrown straight up, reaches a maximum height, then falls to its initial height. Make a
statement about the direction of the velocity and acceleration as the ball is going up.
A) Both its velocity and its acceleration point upward.
B) Its velocity points upward and its acceleration points downward.
C) Its velocity points downward and its acceleration points upward.
D) Both its velocity and its acceleration points downward.
Remember that the acceleration due to gravity is -9.8 m/s2—its direction is downward. As the
ball is thrown upward, once it leaves the thrower’s hand the only thing influencing its motion
(unless otherwise stated—we usually ignore air resistance), is this acceleration due to gravity.
This will cause the ball to decelerate—that is, its velocity will decrease—until it comes to a
stop at its maximum height. However, while the velocity has changed, the acceleration has
not—it remains an acceleration of -9.8 m/s2. After the ball comes to a stop, the negative
acceleration will then cause the ball to reverse direction and increase in velocity in a negative
direction. As it reaches the point where it was thrown, the magnitude (value) of its negative
velocity will equal the magnitude of the positive velocity with which it was thrown. Again,
throughout the whole flight, while the velocity changes, the acceleration does not. So, the
correct answer is, as the ball is going up, the velocity points upwards (in a positive direction),
while the acceleration points downward (in a negative direction). Answer B
21) A ball is thrown straight up, reaches a maximum height, then falls to its initial height. Make a statement about
the direction of the velocity and acceleration as the ball is coming down.
A) Both its velocity and its acceleration point upward.
B) Its velocity points upward and its acceleration points downward.
7 C) Its velocity points downward and its acceleration points upward.
D) Both its velocity and its acceleration point downward.
Remember that the acceleration due to gravity is -9.8 m/s2—its direction is downward. As the
ball is thrown upward, once it leaves the thrower’s hand the only thing influencing its motion
(unless otherwise stated—we usually ignore air resistance), is this acceleration due to gravity.
This will cause the ball to decelerate—that is, its velocity will decrease—until it comes to a
stop at its maximum height. However, while the velocity has changed, the acceleration has
not—it remains an acceleration of -9.8 m/s2. After the ball comes to a stop, the negative
acceleration will then cause the ball to reverse direction and increase in velocity in a negative
direction. As it reaches the point where it was thrown, the magnitude (value) of its negative
velocity will equal the magnitude of the positive velocity with which it was thrown. Again,
throughout the whole flight, while the velocity changes, the acceleration does not. So, the
correct answer is, as the ball is coming down, both its velocity and acceleration point
downward Answer D
22) Suppose a ball is thrown downward in the absence of air resistance. Make a statement concerning its
acceleration.
A) Its acceleration is constantly decreasing.
B) Its acceleration is constantly increasing.
C) Its acceleration is constant.
D) Its acceleration is zero.
Remember that the acceleration due to gravity is -9.8 m/s2—its direction is downward. As the
ball is thrown downward, once it leaves the thrower’s hand the only thing influencing its
motion is this acceleration due to gravity. This will cause the ball to accelerate, increasing its
velocity in a negative direction. While the velocity increases in a negative direction, the
acceleration due to gravity does not change—it is constant. So, the correct answer is, its
acceleration is constant. Answer C
23) Suppose a skydiver jumps from a high-flying plane. What is her acceleration when she reaches terminal
velocity?
A) It is essentially zero.
B) It is in the upward direction.
C) It is approximately 9.8 m/s2 downward.
D) It is a constant pointing upward.
Remember that the acceleration due to gravity is -9.8 m/s2—its direction is downward.
However, as the sky diver reaches terminal velocity, this means her velocity is no longer
changing. Understand that the acceleration due to gravity is still there—it has not changed.
But now, something has opposed the acceleration due to gravity to the point that this
acceleration is balanced by the acceleration due to an opposing force (air resistance). As these
opposing accelerations are balanced and the sky diver is no longer changing velocity, the
acceleration of the sky diver is essentially zero. Answer A
24) A ball is thrown vertically upward with a speed v. An identical second ball is thrown upward with a speed 2v
(twice as fast). What is the ratio of the maximum height of the second ball to that of the first ball? (How many times
higher does the second ball go than the first ball?)
A) 4:1
B) 2:1
C) 1.7:1
D) 1.4:1
This would be one of those times where the use of the term speed is ambiguous—the problem
8 should really be using the term velocity.
When you have questions that ask you to compare how changing a variable will affect a
resulting variable, the best way to see this is by writing out the formula and then change the
affected variable in the manner describe and visualize what would happen to the resulting
variable. Here we are told that both objects start from an initial position (both will have x0 =
0). They will both have the same final velocity when they reach a maximum height (v for both
will be 0 m/s). They will both be under the same influence of acceleration due to gravity (a for
both will be -9.8 m/s2). And the second ball will have an initial velocity that is twice that of the
first. Finally, we are not given a time. This means that in order to compare the two situations,
we will need to use v2 = v02 + 2a(x-x0). However, as we are comparing how high each ball goes
we will want to isolate x. We know that x0 and v2 will drop out.
−v 02
2
2
v = v 0 + 2a(x − x0 ) → x =
2a
Then, write out this formula, highlighting in some way that you want to know what happens to
x if you double initial velocity—I would write it out like this:
− 2 v 02
→ 4x
2a
I have indicated by drawing a box around the 2 in front of the v0 that the change we are
making is to double the initial velocity. Then I write after that, the affect this has on the final
height. This basically shows, “if I double the initial velocity, because this doubling is squared,
I quadruple the final height.” You can waste large quantities of time trying to reason
something out like this in your head if you don’t have a visual cue in front of you. I highly
recommend this approach. Answer A
x=
25) Ball A is dropped from the top of a building. One second later, ball B is dropped from the same building. As
time progresses, the distance between them
A) increases.
B) remains constant.
C) decreases.
D) cannot be determined from the information given.
The confusing part about this question is that you know by now that acceleration for both
balls is the same—they will both accelerate at the same rate—so, the tempting thing to do
would be to immediately say that as time progresses, the distance between the balls must time
trying to reason this out—simply visualize what must be happening to each ball as each
second. After 1 second, ball A has dropped a certain amount but ball B has not yet moved.
After 2 seconds, ball A has dropped an amount greater than it dropped the first second, while
ball B has now dropped the same amount ball A did the first second. You should quickly
realize that at each second, ball B has dropped to the point where ball A had dropped to the
preceding second, and the distance between the two must be increasing. You can also draw a
very quick diagram that helps you visualize this by labeling A and B with subscripts to show
where they are at each second (see diagram)—this only takes a few seconds and saves a lot of
time trying to reason. Answer A
26) Ball A is dropped from the top of a building. One second later, ball B is dropped from the same building. As
time progresses, the difference in their speeds
A) remains constant.
9 B) decreases.
C) increases.
D) cannot be determined from the information given.
Again, the temptation here might be to waste time trying to just reason this out. Again, avoid
the temptation to answer this in a kneejerk fashion. You will be better off in this case by
simply starting to write down the velocities of each ball after each second to actually see what
is happening to the velocities. This is much easier than you might think. You know that
under the influence of gravity, the velocity of each ball will increase by 9.8 m/s each second.
After the first second, ball A will be at a velocity of -9.8 m/s while that of B will be at 0 m/s.
After two seconds, that of A will be at -19.6 m/s, while that of B will be at -9.8 m/s. After three
seconds, that of A will be at -29.4 m/s, while that of B will be at -19.6 m/s.
A
B
1s
-9.8
0
2s
-19.6
-9.8
3s
-29.4
-19.6
By this point you should see that the difference between the velocities of each ball remains
constant. This is because they are both under the influence of the same acceleration. Once
both balls are falling, as the acceleration is the same for both, their velocities are both
increasing at the same rate so the difference in the velocities will remain the same. Answer A
27) Two objects are thrown from the top of a tall building. One is thrown up, and the other is thrown down, both
with the same initial speed. What are their speeds when they hit the street?
A) The one thrown up is traveling faster.
B) The one thrown down is traveling faster.
C) They are traveling at the same speed.
D) It is impossible to tell because the height of the building is not given.
Consider the motion diagram to the left, illustrating the motion of the object thrown down on
the left and the motion of the object thrown upward on the right. If both are thrown with the
same initial velocity, as the object thrown upward decelerates and then comes to a stop, and
then descends again, as this object passes the point where it was thrown upward, its downward
velocity will now have the same magnitude (value) as it did when it was thrown upward. This
means it should now be traveling downward with the same velocity as the initial velocity of the
object that was thrown downward. Therefore, both objects should have the same speed when
they hit the street. Answer C
28) A brick is dropped from the top of a building. A second brick is thrown straight down from the same building.
They are released at the same time. Neglect air resistance. Compare the accelerations of the two bricks.
A) The first brick accelerates faster.
B) The second brick accelerates faster.
C) The two bricks accelerate at the same rate.
D) It is impossible to determine from the information given.
The question is not asking you to compare speeds or velocities. Once each brick leaves the hand, the only thing affecting each is the acceleration due to gravity. Just because one brick is thrown and one is dropped, this does not make the acceleration due to gravity different for one brick than for the other. Once the bricks leave the hand the acceleration is the same for both. Answer C
For questions 29-40:
Make sure you understand the following about position-time graphs.
10 -change in position ( Δx ) is the “rise” of the position-time curve
-change in time ( Δt ) is the “run” of the position-time curve
-therefore, the slope of a position-time graph represents the velocity
rise Δx
slope =
=
=v
run Δt
-because the slope of a linear curve does not change, a linear curve on a positiontime graph represents a constant velocity.
-a horizontal line has a slope of 0 so velocity would be 0 m/s
-a line with a positive slope indicates a constant non-zero
velocity in the positive direction
-a line with a negative slope indicates a constant non-zero
velocity in the negative direction
-any straight line on a position-time graph would also therefore
represent an acceleration of 0
-this type of graph models the equation x = x0 + vt
-because the slope of a non-linear curve does change, a non-linear curve
on a position-time graph represents a non-constant or accelerating
velocity
-the non-linear shape of a curve representing a constant acceleration
will be parabolic and models the equation x = x0 + vot + ½ at2
-because the slope changes, velocity is also changing—however, the
instantaneous velocity can still be estimated because it is the
slope of the tangent to the curve and can be estimated from the
graph
-for a curve representing a positive velocity:
-if the slopes of tangents increase the motion is called
increasing positive velocity (positive velocity and
positive acceleration)
-if the slopes of tangents decrease the motion is called
decreasing positive velocity (positive velocity and
negative acceleration)
-for a curve representing a negative velocity:
-if the slopes of tangents increase the motion is called
increasing negative velocity (negative velocity and
negative acceleration)
-if the slopes of tangents decrease the motion is called
decreasing negative velocity (negative velocity and
positive acceleration)
Make sure you understand the following about velocity-time graphs.
-change in velocity ( Δv ) is the “rise” of the velocity-time curve
-change in time ( Δt ) is the “run” of the velocity-time curve
-therefore, the slope of a velocity-time graph represents the acceleration
rise Δv
slope =
=
=a
run Δt
-because the slope of a linear curve does not change, a linear curve on a velocity 11 time graph represents a constant acceleration.
-a horizontal line has a slope of 0 so acceleration would be 0 m/s
-a line with a positive slope indicates a constant non-zero
acceleration in the positive direction
-a line with a negative slope indicates a constant non-zero
acceleration in the negative direction
-this type of graph models the equation v = v0 + at
-in general, we will only deal with constant acceleration—however, you
should be able to recognize on a velocity time graph that a
non-linear graph represents a non-constant acceleration and
whether it is a positive or negative acceleration and whether it is
increasing or decreasing acceleration by the direction of the slopes of
tangents in fashion similar to that for velocity on a position time graph.
Make sure you understand the following about acceleration-time graphs.
As we mostly deal with situations of constant acceleration, it should occur to you
that the only curve we really need to be concerned about on an acceleration time
graph is one of a horizontal straight line, which would represent a constant
acceleration.
29) An object is moving with constant non-zero velocity in the +x axis. The position versus time graph of this
object is
A) a horizontal straight line.
B) a vertical straight line.
C) a straight line making an angle with the time axis.
D) a parabolic curve.
As the slope of a line on a position time graph represents velocity, “constant velocity” tells you
that the line of the graph must be linear rather than curved. “Non-zero velocity” tells you that
the line must have a non-zero slope. Therefore, the correct answer must be “a straight line
making an angle with the time axis.” Answer C
30) An object is moving with constant non-zero acceleration in the +x axis. The position versus time graph of this
object is
A) a horizontal straight line.
B) a vertical straight line.
C) a straight line making an angle with the time axis.
D) a parabolic curve.
As the slope of a line on a position time graph represents velocity, “acceleration” tells you that
the line of the graph must be curved (changing slope) rather than linear. “Constant non-zero
acceleration” tells you that the line must be parabolic rather than irregularly shaped.
Therefore, the correct answer must be “a parabolic curve.” Answer D
31) An object is moving with constant non-zero velocity in the +x axis. The velocity versus time graph of this
object is
A) a horizontal straight line.
B) a vertical straight line.
C) a straight line making an angle with the time axis.
D) a parabolic curve.
As the slope of the line on a velocity time graph represents acceleration, “constant velocity”—
12 meaning no acceleration—tells you that the line of the graph must be horizontal (0 slope).
“Non-zero” tells you that the line must not be a horizontal line that represents 0 m/s (which is
not necessary to answer this question. Answer A
32) An object is moving with constant non-zero acceleration in the +x axis. The velocity versus time graph of this
object is
A) a horizontal straight line.
B) a vertical straight line.
C) a straight line making an angle with the time axis.
D) a parabolic curve.
As the slope of a line on a velocity time graph represents acceleration, “constant acceleration”
tells you that the line of the graph must be linear rather than curved. “Non-zero” tells you
that the line must have a non-zero slope. Therefore, the correct answer is “a straight line
making an angle with the time axis.” Answer C
33) The slope of a position versus time graph gives
A) position.
B) velocity.
C) acceleration.
D) displacement.
As stated above, the slope of a position time graph represents velocity. Answer B
34) The slope of a velocity versus time graph gives
A) acceleration.
B) position.
C) velocity.
D) displacement.
As stated above, the slope of a velocity time graph represents acceleration. Answer A
35) The area under a curve in an acceleration versus time graph gives
A) acceleration.
B) velocity.
C) displacement.
D) position.
We did not consider this information in class, but with a little thought you should be able to
reason out the answer. The area of any surface is the multiplication of the width times the
length. So, the “area under a curve” on a graph literally represents the area bounded by the
axes and the curve itself, and can be found by multiplying the values of units represented on
each axis. Therefore, as all of our acceleration time graphs will be horizontal straight lines,
representing a constant acceleration, if we multiply whatever the value of the constant
acceleration is on the y axis (lets say 10 m/s2) times the time interval indicated on the x axis
(lets say 5 seconds) we get:
10 m
x 5 s = 50 m / s
s2
That is, the area under the curve of an acceleration time graph represents the final velocity.
Answer B
36) The area under a curve in a velocity versus time graph gives
A) acceleration.
B) velocity.
13 C) displacement.
D) position.
We did not consider this information in class, but with a little thought you should be able to
reason out the answer. The area of any surface is the multiplication of the width times the
length. So, the “area under a curve” on a graph literally represents the area bounded by the
axes and the curve itself, and can be found by multiplying the values of units represented on
each axis. Therefore, on a velocity time graph, if we multiply whatever the value of the
velocity is on the y axis (lets say 10 m/s) times the time interval indicated on the x axis (lets say
5 seconds) we get:
10 m
x 5 s = 50 m
s
That is, the area under the curve of an velocity time graph represents the displacement.
Answer C
37) If the position versus time graph of an object is a horizontal line, the object is
A) moving with constant non-zero speed.
B) moving with infinite speed.
C) moving with constant non-zero acceleration.
D) at rest.
As the slope of a line on a position time graph represents velocity, if the line of the graph is
horizontal this means that the velocity is 0 m/s, and so the object is at rest. Answer D
38) If the position versus time graph of an object is a vertical line, the object is
A) moving with constant non-zero speed.
B) moving with constant non-zero acceleration.
C) at rest.
D) moving with infinite speed.
As the slope of a line on a position time graph represents velocity, if the line of the graph is
vertical this means that the denominator of the velocity formula is 0 and the calculation is
undefined or “infinite.” More accurately, it means that the object is at all positions at the
same instant. Answer D
39) If the velocity versus time graph of an object is a horizontal line, the object is
A) moving with constant non-zero speed.
B) moving with constant non-zero acceleration.
C) at rest.
D) moving with infinite speed.
As the slope of a line on a velocity time graph represents acceleration, if the line of the graph
is a horizontal line this means that the slope, and so the acceleration, is 0. This means that the
object must be moving at a constant speed. Answer A
40) If the velocity versus time graph of an object is a straight line making an angle of 30 degrees with the time axis,
the object is
A) moving with constant non-zero speed.
B) moving with constant non-zero acceleration.
C) at rest.
D) moving with infinite speed.
As the slope of a line on a velocity time graph represents acceleration, if the line of the graph
is a straight line at an angle to the time axis, this means that the slope, and therefore the
acceleration, must be non-zero. Answer B
14 Quantitative Problems
A couple of rules—if you are given certain information you are expected to solve that
problem using the information given—you cannot perform two calculations to avoid
getting around having to use a certain formula when you could have solved the problem
using the formula you did not want to use—in particular—if you can solve a problem using
v2 =v02 + 2a(x-x0), you have to do this instead of using two other calculations.
Also, in a multipart problem that follows calculations involving the same motion or set of
motions, you cannot use a value from a later part of the problem to solve an earlier part—
you do not have “access” to a particular value until you “acquire” it.
To get full credit, for each quantitative problem I need to see a formula, substitution into
the formula and the calculation.
1) An object moves 15.0 m north and then 11.0 m south. Find both the distance traveled and the magnitude of the
displacement vector.
We are given that there is an initial displacement Δx1 = 15.0 m, and then a second
displacement of Δx2 = -11.0 m. Remember that the displacement is a vector quantity that has
both magnitude and direction, and we indicate direction with a positive (positive direction-North) or negative (negative direction—South) sign. However, distance is the amount of
travel without regard to the direction, so to find the distance, we would just add the absolute
values of the displacements to get the total distance. On the other hand, to get the magnitude
(value) of the displacements we add the values of the individual displacements, including the
sign of each displacement, to get the total displacment
d = Δx1 + Δx2 = 15.0 m + 11.0 = 26.0m
Δx = Δx1 + Δx2 = 15.0 m + (−11.0m) = 4.0m
(Note—in class we solved this using initial and final positions of the displacements rather
than just using the displacements—either way is valid—you do need to understand how initial
and final positions relate to the displacements, but the way is easier to show. We also used a
motion diagram—you did not have to. I just needed to see the above formulas, substitution
into the formulas and the calculations.
2) A boat can move at 30 km/h in still water. How long will it take to move 12 km upstream in a river flowing 6.0
km/h?
You are asked to determine a time for a motion and you are given a velocity and a
displacement. You can simply use the basic average velocity formula to fit these given values.
Δx
Δx
You know v =
, but you want time so you rearrange the formula to isolate time: t =
.
t
v
You are given that Δx = 12 km. However, you are given that while the boat can move at 30
km/h in still water, it will be moving against a current of 6 km/h. We could say that the boats
velocity is positive 30 km/h and the current’s velocity is -6 km/h, and sum these velocities to get
an overall velocity. Then plug these values into our formula. This could be shown as:
15 t=
Δx
v
=
Δx
12 km
=
= .5h
v1 + v2 30 km / h + (−6 km / h)
3) 55 mi/h is how many m/s? (1 mi = 1609 m.)
With a conversion, always start by writing down the given value—in this case, 55 mi/h, with
mi in the numerator and h in the denominator. Then think—I want to create a set of
conversion factors that will cancel mi but create m, and will cancel h but create s. I always
start by converting what is in the numerator first. So, for the first factor I have been given a
direct conversion from miles to meters. I want to cancel miles so I put this in the denominator
and, meters in the numerator to create it. I have been given 1 mile = 1609 m so add these
numbers in. Then, for the second conversion factor we want to cancel h, so we put this in the
numerator of the conversion factor and seconds in the denominator to create it. We know 1 h
equals 3600 s so we add these values in. We complete the calculations. There should be 2 sig
figs in the answer. Answer 25m/s
55 mi 1609 m
1h
x
x
= 25 m / s
h
1 mi
3600 s
4) What must be your average speed in order to travel 350 km in 5.15 h?
This is a simple application of the average speed formula. Write out the formula, plug the
values in and solve. There should be 2 sig figs in the result (2 sig figs in 350 km). Answer 68
km/h
s=
d 350 km
=
= 68 km / h
t
5.15 h
5) A runner ran the marathon (approximately 42.0 km) in 2 hours and 57 min. What is the average speed of the
runner in m/s?
This is a simple application of the average speed formula (to find the average speed) and a
unit conversion. The only complication in this case is that we have two different time units to
begin with, hours and minutes, so we might want to just start with our beginning value having
units of minutes—2 hours and 57 minutes is 177 minutes. We don’t really need to solve the
average speed to begin with, we can just include it as the beginning value of the conversion—
42.0 km per 177 minutes. That is, we would just start with 42.0 km in the numerator of the
beginning value and 177 minutes in the denominator. If fact, lets make things even simpler.
We know that 1 km is 1000 m—by this point we can do this conversion in our head so we
shouldn’t really even need to show it—lets just start with 42,000 m in the numerator. But wait,
we can make things even simpler. We know there are 60 s in a minute. Because this is also
such a simple conversion, we could calculate the number of seconds on the calculator (177
minutes x 60 s = 10620 s) and just start with this value and perform just a single calculation.
2 h would be considered an exact number in this problem but 57 min is likely an estimate so
we would want to limit the result to 2 sig figs (if you limited this to 3 sig figs that’s okay, too.
42000m
= 4.0m / s or
10620s
16 Of course, if you are more comfortable, you do not have to take the above short cuts and could
show the full conversion as:
42.0 km 1000m 1min
x
x
= 4.0 m / s
177 min 1 km
60s
Answer, 4.0 m/s
6) A car travels 90 km/h. How long does it take for it to travel 400 km?
Δx
Because we want time
t
Δx
instead of average velocity, we rearrange the formula to isolate t: t =
. Then plug in the
v
given values and solve. There is only 1 sig fig. Answer 4 h
This is an application of the basic average velocity formula v =
t=
Δx 400 km
=
= 4h
v 90km / h
7) A ly (light year) is the distance that light travels in one year. The speed of light is 3.00 × 108 m/s. How many
miles are there in a ly? (1 mi = 1609 m, 1 yr = 365 d.)
With a conversion, always start by writing down the given value—in this case, we actually
have to calculate the beginning value. We are told that light travels 3.00 x 108 m each second,
but we need to know how far light travels in 1 year, so we would need to multiply 3.00 x 108
m/s x the number of seconds in a year to get a beginning value. We would actually do this as
part of our overall calculations knowing that the number of seconds in a year is (60 x 60 x 24
x 365). This then leaves us with a number of meters which we then must convert to miles.
Then think—I want to create a conversion factor that will cancel m but create mi. To cancel
m I will put m in the denominator of the conversion factor; to create mi I will put mi in the
numerator of the conversion factor. I have been given that 1 mi = 1609 m so add these
numbers in the appropriate spots and complete the calculation. There should be 3 sig figs.
Answer 5.88 x 1012 mi
3.00 x 108 m
s
x (60 x 60 x 24 x 365) s x
1 mi
= 5.88 x 1012 mi
1609 m
8) If you are driving 72 km/h along a straight road and you look to the side for 4.0 s, how far do you travel during
this inattentive period?
This problem is asking you to find a displacement for a motion that represents constant
velocity. Even with this simple of a problem, I would make it a habit to write down our basic
problem solving. However, what is in the box is what I need to see.
17 Given
Need
Formula
x0 = 0m
x
x = x0 + vt
Solve
x = (20m / s )(4.0 s ) = 80m
72 km / h
= 20m / s
3.6
t = 4.0 s
v=
9) If you run a complete loop around an outdoor track (400 m), in 100 s, your average velocity is
This is a basic application of the average velocity formula. However, it is also very tricky.
Consider that while the distance you have traveled is 400 m, the displacement is actually 0 m
because your initial and final position are the same—therefore:
Δx 0 m
=
= 0 m/s
t
100s
Your average speed may be 4 m/s but your average velocity is 0 m/s. Answer 0 m/s
v=
10) A polar bear starts at the North Pole. It travels 1.0 km south, then 1.0 km east, then 1.0 km north, then 1.0 km
west to return to its starting point. This trip takes 45 min. What was the bear's average speed?
d
and that distance is the total amount
t
traveled, not the displacement (which would be 0 m in this case). You can see that the total
distance traveled is 4.0 km (you do not need to show that calculation). Just plug the values
into the formula and solve (the answer should be in km/h—just use .75 h). The answer should
have 2 sig figs. Answer 5.3 km/h
Remember that the definition of average speed is s =
s=
d 4.0km
=
= 5.3m / s
t
.75h
11) A polar bear starts at the North Pole. It travels 1.0 km south, then 1.0 km east, then 1.0 km north, then 1.0 km
west to return to its starting point. This trip takes 45 min. What was the bear's average velocity?
Δx
and that displacement is the
t
difference between the final and starting positions. As these positions are the same the
displacement is 0 m. Just plug the values in the formula and solve (the answer should be in
km/h—just use .75 h). Answer 0 km/h
Remember that the definition of average velocity is v =
v=
Δx 0 km
=
= 0 km / h
t
.75 h
12) You are driving home on a weekend from school at 55 mi/h for 110 miles. It then starts to snow and you slow
to 35 mi/h. You arrive home after driving 4 hours and 15 minutes. How far is your hometown from school?
Strategy tip—Working backwards
Many times you will find yourself at a loss as to how to begin a problem, or how to “see your
way through” a problem. In those circumstances you will often spend a great deal of time
18 trying to reason your way through the problem to determine the steps you need to go through.
Spending this amount of time is not a viable option on an AP test. In many cases, when it
appears that you are going to struggle determining the right path to solving the problem, you
should “work backward.” This is how I solve most of the problems I attempt. Using this
problem as an example, as you read through the problem you realize there are two separate
motions combined to complete an overall motion.
Ask—what is it that the problem is asking me to find:
I need the total displacement: Δx
(1a)Ask—do I know a formula that would help me find total displacement:
Based on the reading of the problem, it would be the sum of the individual
displacements.
(1b) Write this formula down and then inspect each component of the formula.
Δx = Δx1 + Δx2
Ask—do I know what the first component is already:
Yes, I know from the information given that the first displacement,
Δx1 = 110 mi
(In the beginning, you should have write this down as part of given
information—but, as your skills grow, you should be able to gather this
information in your head as you work backward through the problem)
Ask—do I know what the second component is already:
No, I need to find a way to calculate this
(2a) Ask—do I know a formula that could determine Δx2 given the information
in the problem:
I could rearrange the average velocity formula to isolate displacement for the second
component of the motion:
Δx
v2 = 2 → Δx2 = v2t2
t2
(ultimately you would want to be able to just write down the equation for the isolated
variable without having to show the isolation)
(2b) Inspect each component of this next equation:
Ask—do I have the velocity of the second component of motion:
19 Yes, I know from the given information that v2 = 35 mi / h
Ask—do I have the time interval of the second component of motion:
No, I need to find a way to calculate this
(3a) Ask—do I know a formula that could determine t2 given the information of the problem:
I have been given the total time of the motion—I could subtract the time from the
first part of the motion from the total time to get the time for the second part of the
motion: t2 = t − t1
(3b) Inspect each component of this next equation:
Ask—do I have the total time:
Yes, In know from the given information that t = 4.25 h
Ask—do I have the time interval of the first component of the motion:
No, I need to find a way to calculate this
(4a) Ask—do I know a formula that could determine the time interval of the first component
of the motion:
I could rearrange the average velocity formula for the first component of the motion
to isolate time:
Δx
Δx
v1 = 1 → t1 = 1
t1
v1
(4b) Inspect each component of this next equation:
Ask—do I have Δx1 :
Yes, I know from the given information that Δx1 = 110 mi
Ask—do I have v1:
Yes, I know from the given information that v1 = 55 mi/hr
I finally have all of the formulas and information I need to solve the problem. Obviously you
do not write all of these questions and answers down—most of it is an internal dialog you have
with yourself. But see the pattern of activity here as outlined by the above (a) and (b) steps—
the whole strategy is just a series of
20 Determine the needed quantity.
(a) Do I know a formula that could determine the quantity I need?
(Yes, you do—but, you may have to think a little sometimes to
create it)
(b) Inspect the components of the formula—if you have the information, great!
If you do not, go back to (a) and repeat.
You can either do this in conjunction with writing down in the given, need, formula, solve
format as we have discussed:
or, you can accomplish without writing the given information down as follows:
Obviously, you do not need to draw the arrows in (but you could if it helped you). The
advantage to writing everything down in a fashion that you cannot forget anything makes it
easier to follow the path to the solution whether you are working in a backward or forward
direction. The disadvantage is that it takes more time—while trying to write everything down
in the beginning, work toward the goal of making most of the given information part of your
internal dialog.
13) A motorist travels 160 km at 80 km/h and 160 km at 100 km/h. What is the average speed of the motorist for
this trip?
As with problem 12, the overall motion contains two separate motions. I would start by
writing down the given and needed information:
Given1
Given2
Need
Δx1 = 160 km Δx2 = 160 km
v1 = 80km / h
v
v2 = 100km / h
21 Writing this information down helps you realize that you need the velocity for the whole trip,
and you can’t just add the two velocities together and divide by 2 to get the average velocity.
(a) Ask—do I know a formula that would help me find average velocity for the whole
trip:
I could just use the average velocity formula.
(b) Write this formula down and inspect the components: v =
Ask—do I have Δx :
Δx
t
Technically no, but I know it will be very easy to find
adding the two component displacements which I have
already been given so I will just add this into my beginning formula:
Δx Δx1 + Δx2
v=
=
t
t
Ask—do I have t for the total trip:
No, but I can easily write a formula to solve for t so I might as
well go ahead and include this as part of my beginning formula too—
as t = t1 + t2, my beginning formula becomes:
Δx Δx1 + Δx2
v=
=
t
t1 + t2
Ask—do I have t1:
No, I need to find a way to calculate for this
(a) Ask—can do I know a formula that could determine t1:
I can rearrange the average velocity formula to solve for t1. In fact, we can
use the same strategy for t2 so we might as well go ahead deal with this now.
(b) Write the formulas down and inspect the components of the formula:
Δx
Δx
t1 = 1
t2 = 2
v1
v2
Ask—do I have the displacements: yes— Δx1 = 160 km Δx2 = 160 km
Ask—do I have the velocities: yes— v1 = 80km / h v2 = 100km / h
As we finally have all of the needed formulas and information we can go ahead and
solve. Again, most of this should be internal dialog occurring in your head—what I
should see as you show your work for the problem is this:
22 Or, if you just show the work and not the given:
Again, the arrows would be only if you needed to help you visualize.
14) A motorist travels for 3.0 h at 80 km/h and 2.0 h at 100 km/h. What is her average speed for the trip?
If you have not done so, go back and review the problem solving strategy outlined in 12 and
13—lets just apply it here without writing everything down. Write down the given information,
determine what you need, and recognize that the initial equation needs to be one that identifies
average velocity. Inspect the average velocity formula to recognize that we don’t have total
displacement, the individual displacements, or total time (but in contrast to the last problem,
we are given the individual times in this problem and so can easily find total time). This
means we will have to create additional formulas to find the individual displacements. We
would do this by rearranging the velocity formula to isolate displacement (see below), and plug
the given information into determine the individual displacements. Then we would plug these
back into our original equations. The answer will have 1 sig fig. Answer 90 km/h
or, without given information:
v=
Δx Δx1 + Δx2 240km + 200km
=
=
= 90km / h
t
t1 + t2
3.0h + 2.0h
15) An airplane travels at 300 mi/h south for 2.00 h and then at 250 mi/h north for 750 miles. What is the average
speed for the trip? (Note—the question asks for average speed—however, because we are more
used to the average velocity formula we are going to allow ourselves to be inconsistent here
and say average velocity—recognizing that if we were being consistent we would have to
23 consider the displacement of the whole trip from where it began to where it ended.)
Again, following our strategy, write down the given information and needed quantity—the trip
consists of two motions and we need average velocity (speed) for the trip. Then write down the
average velocity formula and inspect it, recognizing that we have the velocity and time for the
first component of the motion (v1 and t1) and so will need to calculate Δx1 , and recognizing
that we have displacement and velocity for the second component of the motion ( Δx2 and v2 )
and so will need to calculate t2. That is, we will need to create additional formulas to find Δx1
and t2 in order to solve our beginning equation. The answer will have 1 sigfig. Answer 270
mi/h
or, without the given information:
16) In a 400-m relay race the anchorman (the person who runs the last 100 m) for team A can run 100 m in 9.8 s.
His rival, the anchorman for team B, can cover 100 m in 10.1 s. What is the largest lead the team B runner can have
when the team A runner starts the final leg of the race, in order that the team A runner not lose the race?
As we read through this problem it appears to be very complicated and it may or may
not be helpful to write down our given information because it is not entirely clear in
this problem just exactly what our given information is. But we can still work through
this problem in a backward fashion by asking what it is that we are trying to find.
Because we know that runner A can run 100 m in 9.8 seconds, to find the lead that B
would have to have that would result in a tie we could reason that we would need to
know how far runner B could run in 9.8 s and say that if runner B was at a point
where he had that far to run when runner A started, they would end up in a tie. The
lead runner B would need then would be 100 m – (the distance B could run in 9.8 s)
when runner A started. This gives us our beginning equation—it’s okay to create an
equation that is somewhat descriptive.
lead = 100m − (distance B can run in 9.8s)
Then, as is our strategy, inspect the formula to see if you have the information needed
to solve the formula, and recognize that you need to create a calculation to find
24 (distance B can run in 9.8s)—lets call this Δx . You don’t have this information, but
you then recognize that you can determine this displacement by rearranging the basic
average velocity formula:
Δx = v Bt
Then we inspect this formula and see that we already know time (t = 9.8s). Then
realize that we don’t have v B , so we need to create a formula that determines this.
This is the velocity of runner B and we have been given that he can run 100 m in 10.1 s,
so we can just simply plug these numbers in to get runner B’s velocity— (do not be
confused—the 100 m in 10.1 s is runner B’s velocity—at this velocity, we are trying to
see how far he can go in 9.8s). The result will have 2 sig figs. Answer 3.0 m
lead = 100m − (Δx)= 100m - 97.0m = 3.0m
Δx = v Bt = (9.90m / s)(9.8s) = 97.0m
vB =
Δx B 100m
=
= 9.90m / s
tB
10.1s
17) A car decelerates uniformly and comes to a stop after 10 s. The car's average velocity during
deceleration was 50 km/h. What was the car's deceleration while slowing down?
Gather the given and needed information (see below—remember to convert km/h to
m/s). We are asked for the car’s deceleration, which is the same thing in this case as
asking for the acceleration (it will be negative). As we are given the final velocity and
time and need acceleration it would seem that rearranging the velocity with constant
acceleration formula (v = v0 + at) to isolate a would be a good choice, except that we
don’t have the initial velocity. But, we have been given information that can be used to
find it. Although we do not mention it frequently, we know that for constantly
accelerating motion, the average velocity is the sum of the initial and final velocities
divided by 2. As we know the average velocity and final velocity, we can solve for the
initial velocity. This would allow us to solve the velocity with constant acceleration
formula. The result would have 1 sig fig. Working backwards:
25 or, without the given information:
v = v0 + at → a =
v=
−v0 −2v −(2)(13.89m / s)
=
=
= −3m / s 2
t
t
10s
(v + vo )
→ v0 = 2v
2
Answer -3m/s2
18) An airplane increases its speed from 100 m/s to 160 m/s, at the average rate of 15 m/s2. How much time does it
take for the complete increase in speed?
Gather the given and needed information (see below). As you have v0, v, and a, and you need
time, a logical choice for a formula would be to rearrange the velocity with constant
acceleration formula to isolate t. There is no need to work backwards here, you have all of the
information needed to solve. There should be 1 sig fig in the result. Answer 4 s
Or, without the given information:
v = v0 + at → t =
v − v0 160m / s − 100m / s
=
= 4s
a
15m / s 2
19) A car traveling 60 km/h accelerates at the rate of 2.0 m/s2. How much time is required for the car to reach a
speed of 90 km/h?
Gather the given and needed information (see below). As you have v0, v, (make sure you
convert km/h to m/s) and a, and you need time, a logical choice for a formula would be to
rearrange the velocity with constant acceleration formula to isolate t. . There is no need to
work backwards here, you have all of the information needed to solve. There should be 1 sig
fig in the result. Answer 4 s
Or, without the given information:
v = v0 + at → t =
v − v0 25m / s − 16.67m / s
=
= 4s
a
2.0m / s 2
20) A cart starts from rest and accelerates at 4.0 m/s2 for 5.0 s, then maintain that velocity for 10 s, and then
decelerates at the rate of 2.0 m/s2 for 4.0 s. What is the final speed of the car?
26 In reading through the problem you recognize there are three separate motions. The question
asks for the final speed of the car, and you should recognize that the second motion where the
car maintains a constant velocity will have no affect on the final velocity so we can ignore it.
Also, lets not write down the given information for this problem and just reason through it by
working backwards. You want the final velocity of the car, and for the last motion you are
given a and t, so a good formula would be velocity with constant acceleration (v2f = v2i + at).
(Because you have velocities of two separate motions you might want to separate these in some
way with additional subscripts—for example v2f for final velocity 2.) Writing this formula
down and inspecting it, you recognize that you don’t have v2i, so you need to find a way to
calculate this. You are already aware that the initial velocity for this motion is the final
velocity of the first motion. Therefore, you could find final velocity of the first motion as your
next step, and you could do this also using velocity with constant acceleration and the
information given. Make sure you remember that the second acceleration is in the negative
direction.
Or, without the given information:
v2 f = v2i + a2t2 = 20m / s + (−2.0m / s 2 )(4.0s) = 12 m / s
v2i = v1 f = v1i + a1t1 = (4.0m / s 2 )(5.0s) = 20m / s
Answer 12 m/s
21) A car travels at 15 m/s for 10 s. It then speeds up with a constant acceleration of 2.0 m/s2 for 15 s. At the end
of this time, what is its velocity?
As you read through this problem you see that there are two separate motions. However, as
the question asks for the final velocity after the second motion, and as the first motion is a
constant velocity, you would just need to use the constant velocity of the first motion as the
initial velocity of the second motion. With this in mind, just consider the second motion and
gather the given and needed information. We don’t really know x0 for this second motion so
do not include it. There should be 2 sig figs in this answer. Answer 45 m/s
Given
Need
Formula
v0 = 15m / s
v
v = v0 + at
Solve
v = 15m / s + (2.0m / s 2 )(15s) = 45m / s
a = 2.0m / s 2
t = 15 s
Or, without the given information:
v = v0 + at = 15m / s + (2.0m / s 2 )(15s) = 45m / s
27 22) A cart with an initial velocity of 5.0 m/s experiences a constant acceleration of 2.0 m/s2. What is the cart's
displacement during the first 6.0 s of its motion?
This one seems easy enough to do without writing everything down—lets just work backwards
through it. You are told you want displacement so you will want an equation with “x” in it.
You are given time so you will not need the velocity with constant acceleration/no time
formula. You have been given initial velocity, acceleration and time, but not final velocity, so
you will need the displacement with constant acceleration/initial time equation. As you write
this down and inspect it you see that x0 drops out, and you have all of the other information
you need. The result should have 2 sig figs. Answer 66 m
x = x0 + v0t + 12 at 2 = (5.0m / s)(6.0s) + 12 (2.0m / s 2 )(6s)2 = 66m
23) A bullet moving horizontally to the right (+x direction) with a speed of 500 m/s strikes a sandbag and penetrates
a distance of 10.0 cm. What is the average acceleration, in m/s2, of the bullet?
This one seems easy enough to do without writing everything down—lets just work backwards
through it. You are told you want acceleration so you will want an equation with “a” in it.
You are not given time so you will need the velocity with constant acceleration/no time
formula. You will need to rearrange to isolate at which time you see that v2 and x0 drop out.
At this point you see that you have all of the remaining required information so plug it in and
solve. The result should have only 1 sig fig. Answer -1x106 m/s2
v 2 = v02 + 2a(x − x0 ) → a =
−v02 −(500m / s)2
=
= −1x106 m / s 2
2x
2(.100m)
24) A jet fighter plane is launched from a catapult on an aircraft carrier. It reaches a speed of 42 m/s at the end of
the catapult, and this requires 2.0 s. Assuming the acceleration is constant, what is the length of the catapult?
This one seems easy enough to do without writing everything down—lets just work backwards
through it. You are told you want the length of a catapult—this would be a displacement so
you will want an equation with “x” in it. You are given time so you will not need the velocity
with constant acceleration/no time formula. You have been given both initial and final
velocity, but not acceleration so you will need the displacement with constant acceleration
formula (the one that calculates the average acceleration using both the initial and final
velocity). As you write this down and inspect it you see that x0 and v0 drop out, and you have
all of the other information you need. The result should have 2 sig figs. Answer 42m
x = x0 + 12 (v + v0 )t = (.5)(42m / s)(2.0s) = 42m
25) A car starting from rest moves with constant acceleration of 2.0 m/s2 for 10 s, then travels with constant speed
for another 10 s, and then finally slows to a stop with constant acceleration of -2.0 m/s2. How far does it travel?
As you read through this problem you realize that there are three separate motions combined
in the overall motion. As you want the overall displacement you could start working
backwards by saying that the overall displacement is the sum of the three displacements (see
the initial equation below). As you inspect the equation you know you will have to create a
calculation for all three displacements. For x1 you have initial velocity, a, and t so you know
you will need to use x1 f = x1i + v1it1 + 12 a1t12 . You have all the information so go ahead and
28 complete this calculation because you will need x1 for the next part.
For the second part of the motion you realize that while the car is moving with a constant
velocity, and you can use x2 = x2i + v2t2 to solve for x2, you haven’t been given v2, which would
be the same as the final velocity for the first part of the motion. So, you also need to calculate
this velocity, which would be easiest to calculate using v1 f = v1i + at . You have the
information so go ahead and calculate this and x2 now.
For the third part of the motion you know the car will have an initial velocity of 20 m/s and a
final velocity of 0 m/s. You have also been given acceleration, but not the time that it took. So,
to find the displacement for the third part of the motion you need to rearrange the velocity
with constant acceleration/no time formula to isolate x. I have just shown the formula
isolated for x (make it a goal to be able to isolate variables in your head). You have the
information you need so plug this in and solve. Then, solve for the total displacement. The
result should have 1 sig fig. Answer 400m
26) A car goes from 40 m/s to 80 m/s in a distance of 200 m. What is its average acceleration?
Working backwards through this problem we see that we need acceleration. We immediately
see that we do not have time and so we must use the velocity with constant acceleration/no
time formula, rearranged to isolate acceleration. As we write this down and inspect it we see
that x0 drops out and we have all of the other information we need. The result should only
have 1 sig fig. Answer 10 m/s
2
v 2 − v02 (80m / s ) − (40m / s)
a=
=
= 10m / s 2
2x
2(200m)
2
27) An object starts from rest and undergoes uniform acceleration. During the first second it travels 5.0 m. How
far will it travel during the third second?
To say that you want the displacement during the third second (call this x2s-3s—“displacement
during second 2 to second 3”) means that you need to know the displacement after the third
second (call this x3s—“displacement after 3s”), and also the displacement after two seconds
(call this x2s—“displacement after 2s”), and subtract this from the displacement after three
seconds—in other words: x2s−3s = x3s − x2s
As you inspect this formula you realize that you will need to find a way to calculate the
displacements after 2 and 3 seconds. You have been given time for each of these calculations,
and can at least speculate that you will need the displacement with constant
29 acceleration/initial velocity formula ( x = x0 + v0t + 12 at 2 ) to solve for each. Then recognize,
to find the displacement at each of these times, for both calculations, the initial position and
initial velocity is 0 m and 0m/s respectively—this helps immensely. All we really need then is
to identify an acceleration. At this point I would go ahead and create equations to solve for
x2s and x3s (see below) so you can visualize that all you have left to do is find an acceleration.
Fortunately, you have been given information about the motion during the first second that
will help you find acceleration. For motion during the first second you are given x0 (which
will drop out), x, v0 (which will also drop out) and time, so you can rearrange the displacement
with constant acceleration/initial velocity formula to isolate acceleration, plug in the above
values and solve (see below).
Now that you have acceleration, you can go back and solve for x2s and x3s , and then solve for
displacement during the third second. The result will have 2 sig figs. Answer 25 m
x2s−3s = x3s − x2s = 45m − 20m = 25m
x3s = x0 + v0t3 + 12 at32 = (.5)(10m / s 2 )(3.0s)2 = 45m
x2s = x0 + v0t2 + 12 at22 = (.5)(10m / s 2 )(2.0s)2 = 20m
x1 = x0 + v0t1 + 12 at12 → a =
2x1 2(5.0m)
=
= 10m / s 2
2
t2
(1s)
28) An object is moving in a straight line with constant acceleration. Initially it is traveling at 16 m/s. Three
seconds later it is traveling at 10 m/s. How far does it move during this time?
Working backwards, you need displacement so you need a formula with x in it. You have been
given time so you do not need to use the velocity with constant acceleration/no time formula.
You have been given an initial and final velocity instead of an acceleration so you need the
displacement with constant acceleration formula. After you write this formula down and
inspect it you see that you have all of the information you need so you plug the values in and
solve. The result should only have 1 sig fig. Answer 40 m
x = x0 + 12 (v + v0 )t = (.5)(16m / s + 10m / s)(3s) = 40m
29) A car starts from rest and accelerates uniformly at 3.0 m/s2. A second car starts from rest 6.0 s later at the same
point and accelerates uniformly at 5.0 m/s2. How long does it take the second car to overtake the first car?
This is a very difficult problem. It employs a strategy that you need to become familiar with.
That strategy is—
If there are two objects in motion and you are asked to compare the motion of one
object with the other (as you are in this problem), it is likely that the question will
ask you to say something about the circumstances when the objects are at equal
points or have traveled for an equal amount of time.
30 Recognize—if two objects have traveled the same displacement and/or
for the same amount of time, we can write equations for the displacement
of each object, and, because x would be the same for both, we could also
set the equations of both, equal to each other. This would allow us to
eliminate a variable (x) during our problem solving, simplifying problem
solving when it would appear that we do not have enough information to
solve a problem.
Using this problem as our example of this strategy—
We ultimately need time, so whatever our equations are for solving this problem, they
are both going to need to contain time. THIS IS CRITICAL—we also know that the
displacement both cars have traveled when the second car overtakes the first car will
be the same—that is, whatever x is for each car, it will end up being the same value.
So, our equations both need to contain x. Finally, we are given an acceleration for
both cars and an initial velocity of 0m/s. So, it should make sense that we could write
a displacement with constant acceleration/initial time formula that describes the
motion of each car:
Car 1
Car 2
xcar1 = x0 + v0t + 12 a1t12
xcar 2 = x0 + v0t + 12 a2t22
Fortunately, x0 and v0 are 0m and 0m/s respectively for both cars and drop out of
the respective formulas. However, we do need to make a minor adjustment for the
amount of time each car is traveling. In most of these problems, the amount of time
each object is traveling will be the same. This would mean that we could just say
“t” in each equation and not worry about t1 or t2. However, in this problem, we
have been told that car 2 leaves 6.0 s after car 1, so car 2 travels for six seconds
less than car 1. Recognize that as we don’t have any other values for t1 or t2, we
would have two more variables than we could solve for if we left these variables
as t1 and t2, so we need to put one of the times in terms of the other time, and this
will eliminate one of these variables. As we want the time it takes car 2 to overtake
car 1, lets put the time for car 1 in terms of the time for car 2. That is, lets say
t1 = t2 + 6s. This means we can modify the above equations to say:
This is unfortunate because it means we will end up having to solve a quadratic
formula to find t2. But, oh well.
But now, we bring the CRITICAL PART back into our strategy—recognize again
that at the point car 2 catches car 1, the “x” for both cars will be the same. This
means that:
1
a (t + 6s)2 will also equal 12 a2t22
2 1 2
31 We can then isolate t2, plug in the values we have been given for a1 and a2, and solve
for t2 (again, unfortunately, using a quadratic formula).
Now, plug values into the quadratic formula and solve for t2. The result should have 2
sig figs.
t2 =
−36 + (36)2 − (4)(−2)(108)
= −2.7s this cannot be a solution − cannot have negative time
2(−2)
−36 − (36)2 − (4)(−2)(108)
t2 =
= 21s this is our solution
2(−2)
What I would need to see as work is what is in the boxes above. Answer 21s
30) A car with good tires on a dry road can decelerate at about 5.0 m/s2 when braking. Suppose a car is initially
traveling at 55 mi/h.
(a.) How much time does it take the car to stop?
(b.) What is the stopping distance?
(a) Working backwards, we need time, we are given and acceleration, and initial and final
velocities. (We will need to convert to m/s—I would do this in the context of the calculation—
see below). This means we should be able to use the velocity with constant acceleration
formula (v = v0 + at). We have all of the information so plug values in and solve. (Remember
that the acceleration is negative). The result will contain 2 sig figs. Answer 4.9s
⎛ 55 mi ⎞ ⎛ 1609m ⎞ ⎛ 1 h ⎞
−⎜
⎟⎜
⎟
⎟⎜
−v0
⎝ h ⎠ ⎝ 1 mi ⎠ ⎝ 3600s ⎠
v = v0 + at → t =
=
= 4.9s
a
−5.0m / s 2
(b)Now that you have time, this second part will be easier. We need displacement so you need
a formula with x. You know x0, v0, v, a, and now t, so the displacement with constant
acceleration formula (x = x0 + ½ (v + v0)t) or displacement with constant acceleration/initial
velocity formula (x = x0 + v0t + ½ at2) would work. (Note—at this point you might recognize
the disadvantage of having added the conversion to m/s into the calculation above—you did
not actually calculate this velocity—I would not write all of this down again for initial velocity,
32 I would just calculate on the calculator and add it in as a know value.) This result should
have 2 sig figs. Answer 60 m
x = x0 + v0t + 12 at 2 = (24.58m / s)(4.9s) + (.5)(−5.0m / s 2 )(4.9s)2 = 60m
31) At the instant a traffic light turns green, a car that has been waiting at the intersection starts ahead with a
constant acceleration of 2.00 m/s2. At that moment a truck traveling with a constant velocity of 15.0 m/s overtakes
and passes the car.
(a.) Calculate the time necessary for the car to reach the truck.
(b.) Calculate the distance beyond the traffic light that the car will pass the truck.
(c.) Determine the speed of the car when it passes the truck.
(a) This is another one of those problems where we have two objects in motion and we are to
compare the motion of one object with the other. Again, it is likely that the question will
ask you to say something about the circumstances when the objects are at equal points or have
traveled for an equal amount of time. Remember, the CRITICAL POINT is that the
displacement both vehicles have traveled when the second vehicle overtakes the first will
be the same—that is, whatever x is for the truck will be x for the car. This problem will not be
as difficult as the previous one as we will not need to solve a quadratic formula—however, the
strategy is the same.
We have been given information about the motion of both the car and the truck so
we should be able to create formulas that would calculate the displacement for each
vehicle at the time the car catches up to the truck. Think—what would writing such a
displacement formula for each vehicle involve.
For the car, we are given x0 (which will drop out), v0 (for which v0t will
drop out, and acceleration. The equation for the car will thus use
the displacement with constant acceleration/initial velocity formula.
For the truck, it is traveling at a constant velocity. We have been given
x0 (which will drop out) and v.
Time for both will be “t”—both vehicles will have been traveling for the
same amount of time when the car catches the truck.
Then, because xcar would equal xtruck, we could set their equations equal to each other
and solve for t. The result should have 3 sig figs. Answer 15.0s
33 Car
xcar = x0 + v0t + 12 acar t
1
2
1
2
Truck
2
xtruck = x0 + vtruck t
acar t 2 = vtruck t
acar t = vtruck
t=
2vtruck 2(15.0m / s)
=
= 15.0s
acar
2.00m / s 2
(b) Now that you have time, and you have previously created displacement equations for
both the car and the truck, you should be able to plug the time you got in (a) into either of
these equations and get the same answer for displacement. I would recommend plugging time
into both to check your answer—if they are not the same, you’ve done something wrong. This
answer should have 3 sig figs. Answer 225 m
Car
xcar = x0 + v0t + 12 acar t 2 = 12 (2.0m / s 2 )(15.0s)2 = 225m
Truck xtruck = x0 + vtruck t = (15.0m / s)(15.0s) = 225m
(c) We could use a variety of formulas to calculate this but the easiest one to use would be
the velocity with constant acceleration formula. You have initial velocity, acceleration and
time for the car—plug these in and solve. The result should have 3 sig figs. Answer
v = v0 + at = (2.0m / s 2 )(15.0s) = 30m / s
32) An object is thrown upward with a speed of 12 m/s on the surface of planet X where the acceleration due to
gravity is 1.5 m/s2. What is the maximum height reached by the object?
Working backward, you want a displacement so the equation will need to have “x” in it. You
do not need time and have not been given time so you will need to use the velocity with
constant acceleration/no time formula and rearrange it to isolate for x. You have been given
x0 = 0m (so this will drop out), v0, v (v = 0 m/s at maximum height—so, this will drop out), and
a (remember that this is in a negative direction). So, after writing the equation down and
inspecting it, you see that you have all of the information. Plug the values in and solve. The
answer should have 2 sig figs. Answer 48m
x=
−v02
−(12m / s)2
=
= 48m
2a 2(−1.5m / s 2 )
33) An object is thrown upward with a speed of 12 m/s on the surface of planet X where the acceleration due to
gravity is 1.5 m/s2. How long does it take for the object to reach the maximum height?
Because this is a separate question, you do not have “access” to the information in the
previous problem—x = 48m is not available to you. Working backward, you want a time so
the equation will need to have “t” in it. You have been given v0, v (v = 0 m/s at maximum
height—so, this will drop out), and a (remember that this is in a negative direction). The
obvious choice for an equation is the velocity with constant acceleration equation, which will
34 need to be rearranged to isolate t (you should be able to just write down the isolation equation).
After writing the equation down and inspecting it, you see that you have all of the information.
Plug the values in and solve. The answer should have 2 sig figs. Answer 8s
t=
v − v0 −(12m / s)
=
=8s
a
−1.5m / s 2
34) An object is thrown upward with a speed of 15 m/s on the surface of planet X where the acceleration due to
gravity is 2.5 m/s2. How long does it take for the object to return to where it is thrown?
Working backwards we want time, so we need a formula that contains “t.” We have been
given x0 (x0 = 0m so this will drop out), v0, and a. But, whether you realize it or not, we have
also been given v. If we ignore air resistance (as we usually do), we realize that the velocity
that the ball returns to the point at which it was thrown should be the same as the velocity with
which it was thrown, just in the opposite direction. (Use the motion diagram at the side to help
you see the truth of this reasoning.) Therefore, v = -15m/s. Given all of this information, we
can now see that the velocity with constant acceleration formula with t isolated will work for
us. We have all of the information—just plug in and solve. The answer should contain 2 sig
figs. Answer 12s
t=
v − v0 −15m / s − 15m / s
=
= 12 s
a
−2.5m / s 2
35) An object is thrown upward with a speed of 14 m/s on the surface of planet X where the acceleration due to
gravity is 3.5 m/s2. What is the speed of the object after 8.0 s?
We want a final velocity (a velocity after 8.0s). Recognize that at the 8 second point in the
motion the velocity might either be positive or negative—you won’t know until you actually
solve for it. In addition to time, you are given initial velocity and acceleration and the obvious
choice for a formula is velocity with constant acceleration. Make sure you are aware that in
order to get the correct sign of the final velocity (positive or negative) the sign of acceleration
has to be correct. The result should have 2 sig figs. Answer -14m/s
v = v0 + at = 19.6 m / s + (−9.8m / s 2 )(3.00s) = −9.80 m / s 2
36) Human reaction time is usually greater than 0.10 s. If your friend holds a ruler between your fingers and
releases it without warning, how far can you expect the ruler to fall before you catch it?
Because we want to know how far the ruler falls, we need a formula that has “x” in it. We are
given a time so we do not need to use the velocity with constant acceleration no time formula.
We also know x0 (=0m, so it will drop out), v0 (=0m/s so a v0t term will drop out), and
acceleration (=-9.8m/s2). We have talked ourselves into using the displacement with constant
acceleration/initial velocity formula. We have all of the information we need. Plug the values
in and solve. The answer should contain 2 sig figs. Answer .049m
x = x0 + v0t + 12 at 2 = (.5)(−9.8m / s 2 )(.10s)2 = .049m
35 37) A ball is thrown upward at a velocity of 19.6 m/s. What is its velocity after 3.00 s?
We want to know a final velocity so our formula needs to have “v” in it. We are also given x0
(=0m, so this would drop out), v0 (= 19.6 m/s), acceleration (-9.8m/s2) and time (3.0s). We can
see that the formula we need to use is the velocity with constant acceleration formula. We
have all the information we need to plug in the values and solve. The answer should contain 3
sig figs. Answer -9.80 m/s2
v = v0 + at = 19.6 m / s + (−9.8m / s 2 )(3.00s) = −9.80 m / s 2
38) A bullet shot straight up returns to its starting point in 10 s. What is the initial speed of the bullet?
We want to know an initial velocity. We are given time (= 10s), and we know acceleration (=
-9.8 m/s2). Logically it would seem that the velocity with constant acceleration formula would
be a good choice (v = v0 + at) but we have not been given an initial velocity. However, we can
remember that for an object that is projected upward with a certain initial velocity, if it falls
back to its original point, the negative velocity it reaches at that point will have a magnitude
that is the same as the velocity with which it was projected upward. That is, final velocity
equals the opposite of the initial velocity, v = -v0 — or, v0 = -v is also true . We can use this to
eliminate a variable in the formula we have chosen. As we want to solve for initial velocity
(v0), lets substitute –v0 for v in the formula (see below). Then, manipulate the formula to
isolate v0 and plug in a and t and solve. The result should contain 1 sig fig. Answer 50 m/s
v = v0 + at
−v0 = v0 + at
2v0 = −at
v0 =
−at −(−9.8m / s 2 )(10s)
=
= 50m / s
2
2
39) A ball is thrown straight up with a speed of 36.0 m/s. How long does it take to return to its starting point?
Working backwards we want time, so we need a formula that contains “t.” We have been
given x0 (x0 = 0m so this will drop out), v0, and a. But, whether you realize it or not, we have
also been given v. If we ignore air resistance (as we usually do), we realize that the velocity
that the ball returns to the point at which it was thrown should be the same as the velocity with
which it was thrown, just in the opposite direction. (Use the motion diagram at the side to help
you see the truth of this reasoning.) Therefore, v = -36.0m/s. Given all of this information, we
can now see that the velocity with constant acceleration formula with t isolated will work for
us. We have all of the information—just plug in and solve. The answer should contain 3 sig
figs. Answer 7.35 s
t=
v − v0 −36m / s − 36m / s
=
= 7.35 s
a
−9.8m / s 2
40) A ball is thrown downward from the top of a building with an initial speed of 25 m/s. It strikes the ground after
2.0 s. How high is the building?
We are asked how tall the building is so we want a displacement—we need “x” in our formula.
However, as we don’t want a negative value in our answer it would be better if we actually let
36 x0 represent the height of the building and let x be our “0” value. So, it would be x that drops
out of such a formula rather than x0. We are given v0 (remember that it is a negative
direction), a ( = -9.8 m/s2) and t, so it would be logical that the displacement with constant
acceleration/initial velocity formula would be used, with x0 isolated. We have all of the values
so plug them in and solve (remember that v0 is -25 m/s). The answer should have 2 sig figs.
Answer 70. m
x0 = x − v0t − 12 at 2 = −(−25m / s)(2.0s) − (.5)(−9.8m / s 2 )(2.0s)2 = 70. m
41)
(a.)
(b.)
(c.)
A ball is thrown straight up with a speed of 30 m/s.
How long does it take the ball to reach the maximum height?
What is the maximum height reached by the ball?
What is its speed after 4.2 s?
(a) This part of the problem requires time so the formula needs to contain “t.” You have been
given an initial velocity (v0 = 30 m/s) and you know that at maximum height final velocity
equals 0 m/s (v = 0 m/s). You also know the acceleration (a = -9.8 m/s2). Therefore, you
should be able to find time by rearranging the velocity with constant acceleration formula to
isolate t. You have all the given information you need so plug in values and solve. The final
answer should contain 1 sig fig. However, as you will likely use this answer in another part of
the problem it is always a good idea in these multipart problems to show both a rounded and
unrounded answer. You can use the unrounded answer in other parts of the problem to help
avoid rounding error. Answer 3 m/s
t=
v − v0
−30m / s
=
= 3.0612s = 3s
a
−9.8m / s 2
(b) This part of the problem is looking for maximum height (displacement) so the formula
must contain “x.” In addition to time (available from part (a)—if you use this you should use
an unrounded value—a reason to a show value both before and after you round to sig figs in
your answer), you again have v0 and v available to you. You could use either the displacement
with constant acceleration or displacement with constant acceleration/initial velocity formula,
or you could use the velocity with constant acceleration/no time formula to find x (this would
reduce rounding error because you are not depending on the calculated value of time from the
previous calculation). x = xo + ½ (v+v0) would be the simplest. You already have all of the
information—plug in and solve. The answer should contain 1 sig fig—however, again, show
both a rounded and unrounded value in case you need this displacement for an additional
calculation. Answer =50 m
x = x0 + 12 ( v + v0 )t = (.5)(30m / s)(3.0612s) = 45.918 m = 50m
(c) This part of the problem requires that you find a final velocity (velocity at 4.2s). Remember
that during the flight of the ball, after 4.2 s the ball may either be still rising or it may be on
its way back down. These means that you have to make sure the signs of your acceleration
and velocity are correct. You know acceleration and you have been given initial velocity so
plugging these values into the velocity with constant acceleration formula should give you the
correct answer. The result should contain 1 sig fig. Answer -10m/s (so it is on its way down).
37 v = v0 + at = 30 m / s + (−9.8m / s 2 )(4.2s) = −11.16 m / s = −10 m / s
42) A foul ball is hit straight up into the air with a speed of 30.0 m/s.
(a.)
Calculate the time required for the ball to rise to its maximum height.
(b.)
Calculate the maximum height reached by the ball.
(c.)
Determine the time at which the ball pass a point 25.0 m above the point of contact between the bat and
ball.
(d.)
Explain why there are two answers to part c.
(a) Calculation is exactly the same as for part a in 41 (except that there are 3 sig figs).
t=
v − v0
−30m / s
=
= 3.06s
a
−9.8m / s 2
(b) Calculation is exactly the same as for part b in 41 (except that there are 3 sig figs).
x = x0 + 12 ( v + v0 )t = (.5)(30m / s)(3.06s) = 45.9m
(c) The third part of this problem requires that you determine a time so you need a “t” in the
formula. You have also been given a final position (x = 25.0 m) and an initial velocity. You
do not have a final velocity so you cannot use velocity with constant acceleration or
displacement with constant acceleration. Also, because it does not have time in it, you cannot
use the velocity with constant acceleration/no time formula. So, you are left with the
displacement with constant acceleration/initial velocity formula, and you would need to
rearrange it to isolate time. Here’s the problem—it won’t be very long before you realize that
you need to solve a quadratic formula to find t—and interestingly, both solutions to the
quadratic formula will be valid solutions because one will be the time for passing 25.0 m on
the way up while the other will be the time for passing the 25.0 m point on the way down.
First, rearrange the equation so that it is in the form of a quadratic equation. Then substitute
in the values for a, v0, and x. Then plug the coefficients into the quadratic formula and accept
both solutions.
x = x0 + v0t + 12 at 2 →
1
2
1
2
at 2 + v0t − x = 0
(−9.80)t 2 + 30.0t − 25.0 = 0
−4.9t 2 + 30.0t − 25.0 = 0
(a = −4.9;b = 30.0;c = −25.0)
t=
−30.0 + (30.0)2 − (4)(−4.9)(−25.0)
= .995s
2(−4.9)
t=
−30.0 − (30.0)2 − (4)(−4.9)(−25.0)
= 5.13s
2(−4.9)
(d) There are two answers in part c because one answer is for when the ball passes 25.0m on
the way up while the other is for when the ball passes 25.0m on the way down.
38 FIGURE 2-1
43) In Fig. 2-1, what is the velocity at t = 1.0 s?
Remember that on a position time graph the slope of the curve represents the velocity and that
a linear curve represents a constant velocity. As the velocity appears to be constant from 0 to 2
seconds (because of the linear graph) we can estimate the slope of this part of the curve to find
the velocity at 1 s. The position appears to change from 0 to 20 m during the first 2.0 seconds.
We can plug these values into the velocity formula. The answer has 1 significant figure.
Answer 10 m/s
v=
Δx 20m
=
= 10m / s
t
2.0s
44) In Fig. 2-1, what is the velocity at t = 2.5 s?
Remember that on a position time graph the slope of the curve represents the velocity and that
a linear curve represents a constant velocity. As the velocity appears to be constant from 2 to 3
seconds (because of the linear graph) we can estimate the slope of this part of the curve to find
the velocity at 2.5 s. The position appears to change from 20 to 40 m during this second. We
can plug these values into the velocity formula. The answer has 1 significant figure. Answer
10 m/s
v=
Δx 20m
=
= 20m / s
t
1.0s
45) In Fig. 2-1, what is the velocity at t = 4.0 s?
Remember that on a position time graph the slope of the curve represents the velocity and that
a linear curve represents a constant velocity. Additionally, from 3.0 to 5.0 s, the line is
horizontal indicating that during this time interval position does not change—the slope is 0.
Therefore, the velocity is 0 m/s. You do not actually need to perform a calculation for this—
you can simply state the on a position time graph, because slope represents velocity, and
because a horizontal line has a slope of 0, a horizontal line represents a constant velocity of 0
39 m/s. Answer 0 m/s
v=
Δx 0m
=
= 0m / s
t
2.0s
46) In Fig. 2-1, what is the velocity at t = 5.5 s?
Remember that on a position time graph the slope of the curve represents the velocity and that
a linear curve represents a constant velocity. As the velocity appears to be constant from 5 to 6
seconds (because of the linear graph) we can estimate the slope of this part of the curve to find
the velocity at 5.5 s. The position appears to change from 40 to 0 m during this second
(remember that displacement is final position minus initial position so displacement is in the
negative direction here). We can plug these values into the velocity formula. The answer has
1 significant figure. Answer -40 m/s
v=
Δx −40m
=
= −40m / s
t
1.0s
47) In Fig. 2-1, what is the average velocity from 0 to 4.0 s?
Remember that on a position time graph the slope of the curve represents the velocity and that
a linear curve represents a constant velocity. As opposed to 43-46 where we were only
considering a single component of the motion, in this problem the time interval covers three
different motions. However, all we have to worry about for the average velocity is the total
displacement and total time interval—we don’t have to worry about adding the three
displacements together and the three time intervals together—we can already “see” this
information on the position time graph. We can plug these values into the average velocity
formula directly and calculate. The answer has 1 sig fig. Answer 10 m/s
48) In Fig. 2-1, what is the average velocity from 0 to 6.0 s?
Remember that on a position time graph the slope of the curve represents the velocity and that
a linear curve represents a constant velocity. As opposed to 43-46 where we were only
considering a single component of the motion, in this problem the time interval covers four
different motions. However, all we have to worry about for the average velocity is the total
displacement and total time interval—we don’t have to worry about adding the four
displacements together and the four time intervals together—we can already “see” this
information on the position time graph. In fact, we should be able to immediately see that
both the beginning and final position are the same—therefore the displacement is 0m, and so
the velocity will be 0m/s. The answer has 1 sig fig. Answer 0 m/s
v=
Δx 0m
=
= 0m / s
t
6s
40 FIGURE 2-2
49) In Fig. 2-2, what is the acceleration at 1.0 s?
Remember that on a velocity time graph the slope of the curve represents the acceleration and
that a linear curve represents a constant acceleration. As the acceleration appears to be
constant from 0 to 2 seconds (because of the linear graph) we can estimate the slope of this
part of the curve to find the acceleration at 1 s. The velocity appears to change from 0 to 20
m/s during the first 2.0 seconds. We can plug these values into the acceleration formula. The
answer has 1 significant figure. Answer 10 m/s2
a=
Δv 20m / s
=
= 10m / s 2
t
2.0s
50) In Fig. 2-2, what is the acceleration at 3.0 s?
Remember that on a velocity time graph the slope of the curve represents the acceleration and
that a linear curve represents a constant acceleration. Additionally, from 2.0 to 4.0 s, the line
is horizontal indicating that during this time interval velocity does not change—the slope is 0.
Therefore, the acceleration is 0 m/s. You do not actually need to perform a calculation for
this—you can simply state the on a velocity time graph, because slope represents acceleration,
and because a horizontal line has a slope of 0, a horizontal line represents a constant
acceleration of 0 m/s. Answer 0 m/s
a=
Δv 0m / s
=
= 0m / s 2
t
2.0s
51) In Fig. 2-2, what is the average acceleration from 0 to 5.0 s?
Remember that on a velocity time graph the slope of the curve represents the acceleration and
that a linear curve represents a constant acceleration. As opposed to 49-50 where we were
only considering a single component of the motion, in this problem the time interval covers
three different motions. However, all we have to worry about for the average acceleration is
the overall change in velocity and overall time interval—we don’t have to worry about
“averaging” the three velocities together and the three time intervals together—we can already
“see” this information on the velocity time graph. We can plug these values into the average
acceleration formula directly and calculate. The answer has 1 sig fig. Answer 10 m/s
41 a=
Δv 10m / s − 0m / s
=
= 2m / s 2
t
5s
52) In Fig. 2-2, what is the average acceleration from 0 to 8.0 s?
Remember that on a velocity time graph the slope of the curve represents the acceleration and
that a linear curve represents a constant acceleration. As opposed to 49-50 where we were
only considering a single component of the motion, in this problem the time interval covers
four different motions. However, all we have to worry about for the average acceleration is
the overall change in velocity and overall time interval—we don’t have to worry about
“averaging” the three velocities together and the three time intervals together—we can already
“see” this information on the velocity time graph. We can plug these values into the average
acceleration formula directly and calculate. The answer has 1 sig fig. Answer 10 m/s
a=
Δv −20m / s − 0m / s
=
= −2.5m / s 2
t
8s
53) In Fig. 2-2, what is the displacement from 0 to 8.0 s?
Remember that the area “under a velocity time graph” is the displacement. So, as the area under this velocity time graphs represents a series of rectangles and triangles, we should be able to figure out this area. Consider the drawing of the velocity time graph to the right which simply splits up the area under the curve into three triangles and one rectangle. For the rectangle A, notice that the velocity on the y axis is 20 m/s and the time interval is 2.0 s. If we find the area of this rectangle we are multiplying the height (20 m/s) times the base (2.0s). (20m / s) x (2.0 s) = 40m Notice that the answer to this calculation is in the units of meters, which is what we want—during the time interval from 2-­‐4 seconds there is 40 m of displacement. Now notice triangles B and C—visualizing inspecting each reveals that they are exactly ½ the area of the rectangle, and added together, they would equal the area of the another whole rectangle. So, the displacement over the first 6 seconds is 40m + 40m = 80m. Finally, triangle D is also ½ the area of the rectangle—again, this represents ½ the area of the rectangle, it represents a displacement of 20 m. However, it represents motion in a negative direction—back toward the origin—so it must be subtracted from the first 80 m of displacement to given a total of 60 m of 42 displacement. The overall calculation for each of the four areas would look like: 20m + 40m + 20m − 20m = 60m Answer 60 m 43