October 6/7, 2015 AP Physics 1 Test (1) Kinematics in One Dimension AK MULTIPLE CHOICE. Choose the one alternative that best completes the statement or answers the question. 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 is either greater than or equal to the distance traveled. B) The displacement is either less than or equal to the distance traveled. C) The displacement can be either greater than, smaller than, or equal to the distance traveled. D) The displacement is always 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 B
2. When is the average velocity of an object equal to the instantaneous velocity? A) always B) only when the velocity is constant C) only when the velocity is increasing at a constant rate D) never Δx x2 − x1
-­‐-­‐the total displacement divided =
Δt t2 − t1
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 B 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) average acceleration. B) average speed. C) instantaneous 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 that plug into the average acceleration equation, a =
1 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 A
4. 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
5. Suppose that a car traveling to the South (-­‐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 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 6. 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 inversely proportional to its surface area. C) The speed of the falling object is proportional to its weight. D) None of the above is true. 2 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 7. 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) Its velocity points upward and its acceleration points downward. B) Its velocity points downward and its acceleration points upward. C) Both its velocity and its acceleration point downward. D) Both its velocity and its acceleration point upward. 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 C
8. 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 two bricks accelerate at the same rate. B) The second brick accelerates faster. C) The first brick accelerates faster. 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 A
9. An object is moving with constant non-­‐zero velocity in the +x axis. The velocity versus time graph of this object is A) a parabolic curve. B) a straight line making an angle with the time axis. C) a horizontal straight line. D) a vertical straight line. As the slope of the line on a velocity time graph represents acceleration, “constant velocity”—
meaning no acceleration—tells you that the line of the graph must be horizontal (0 slope).
3 “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 C
10. Objects A and B both start at rest. They both accelerate at the same rate. However, object A accelerates for three times as long as object B. What is the final speed of object A compared to that of object B? A) twice as fast B) three times as fast C) four times as fast D) nine 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 triple time—I would write it out like this:
v = v0 + a 3 t → 3 v
I have indicated by drawing a box around the 3 in front of the t that the change we are making
is to triple the time. Then I write after that, the affect this has on the velocity. This basically
shows, “if I triple the time, I triple 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
11. Objects A and B both start from rest. They both accelerate at the same rate. However, object A accelerates for three times as long as object B. What is the distance traveled by object A compared to that of object B? A) twice as far B) three times as far C) four times as far D) nine times as far In this case we are given the same information as in 10, 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 triple t. Write out this formula, highlighting in some way that
you want to know what happens to x if you triple time—I would write it out like this:
x = x0 + v0t + 12 a 3 t 2 → 9 x
I have indicated by drawing a box around the 3 in front of the t that the change we are making
is to triple the time. Then I write after that, the affect this has on the displacement. This
basically shows, “if I triple the time, because the tripling is squared, I multiply the
displacement by nine times.” 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 D
4 12. Refer to the following position time graphs-­‐select the one best description for each numbered line.
11.
12.
13.
14.
15.
a. constant positive velocity
b. constant velocity of 0 m/s
c. constant negative velocity
d. increasing positive velocity
e. decreasing positive velocity
f. increasing negative velocity
g. decreasing negative velocity
13. Refer to the following velocity time graphs-­‐select the one best description for each numbered line .
16.
17.
18.
19.
20.
a. constant positive acceleration
f. increasing negative acceleration
b. constant acceleration of 0 m/s g. decreasing negative acceleration
c. constant negative acceleration
d. increasing positive acceleration
e. decreasing positive acceleration
2
11. Parabolic curve on a PT graph indicates changing velocity (acceleration); negative slope indicates negative direction; increasing slopes of tangents to curve indicates increasing velocities—increasing negative velocity. Answer f 12. Straight line on a PT graph indicates constant velocity; horizontal line means 0 slope which means velocity of 0—constant velocity of 0 m/s. Answer b 13. Parabolic curve on a PT graph indicates changing velocity (acceleration); positive slope indicates positive direction; increasing slopes of tangents to curve 5 indicates increasing velocities—increasing positive velocity. Answer d 14. Parabolic curve on a PT graph indicates changing velocity (acceleration); positive slope indicates positive direction; decreasing slopes of tangents to curve indicates decreasing velocities—decreasing positive velocity. Answer e 15. Straight line on a PT graph indicates constant velocity; negative slope indicates a negative velocity—constant negative velocity. Answer c 16. Parabolic curve on a VT graph indicates changing acceleration; positive slope indicates positive direction; increasing slopes of tangents to curve indicates increasing acceleration—increasing positive acceleration. Answer d 17. Straight line on a VT graph indicates constant acceleration; positive slope indicates a positive acceleration—constant positive acceleration. Answer a 18. Straight line on a VT graph indicates constant acceleration; negative slope indicates a negative acceleration—constant negative acceleration. Answer c 19. Parabolic curve on a VT graph indicates changing acceleration; positive slope indicates positive direction; decreasing slopes of tangents to curve indicates decreasing acceleration—decreasing positive acceleration. Answer e 20. Parabolic curve on a VT graph indicates changing acceleration; negative slope indicates negative direction; increasing slopes of tangents to curve indicates increasing acceleration—increasing negative acceleration. Answer f For questions 14 through 18 consider the following position time graph:
14. What is the velocity during segment B of the motion? For 1 point of extra credit show work to support your answer. A) 1.0 m/s B) .67 m/s C) -­‐2.0 m/s D) -­‐.67 m/s The slope of a linear curve on a position time graph represents the velocity—it is a graphical 6 representation of the relationship, v =
Δx
. We can estimate the velocity from the graph by t
determining the rise (representing the displacement) and dividing it by the run (representing the time interval). In this case, the time interval of segment is three seconds, and during that time the position changes from 2 m to 0 m—a displacement of -­‐2 m. So, the velocity is v=
Δx −2m
=
= −.67m / s t
3s
Answer D 15. The velocity is 0 m/s during A) no segments of the motion B) one segment of the motion C) two segments of the motion D) three segments of the motion As the slope of the position time represents the velocity, if the velocity is 0 m/s, the slope must be 0, or a horizontal line. In this case the slope is horizontal during segment A, C, and E. So, the velocity is 0 m/s during three segments of the motion. Answer D 16. What is the velocity during segment F of the motion? For 1 point of extra credit show work to support your answer. A) 1 m/s B) 3 m/s C) 0 m/s D) -­‐3 m/s See discussion for question 14. In this case, the time interval of the given segment is two seconds, and during that time the position changes from -­‐3 m to 3 m—a displacement of 6 m. So, the velocity is v =
Δx 6m
=
= 3m / s t
2s
Answer D 17. What is the average velocity during the first 9 seconds of the motion? For 1 point of extra credit show work to support your answer. A) 0 m/s B) -­‐.33 m/s C) -­‐.67 m/s D) -­‐1 m/s I messed up on this question. In this case the given time interval is nine seconds, and during that time the position changes from 2 m to -­‐3 m—a displacement of -­‐5 m. So the velocity is v=
Δx −5m
=
= −.56m / s When I wrote this question for some reason my brain thought the t
9s
initial position was 0 m instead of -­‐2 m, in which case velocity would have been v=
Δx −3m
=
= −.33m / s . Because of this, I removed this question, although if you showed me t
9s
evidence that you understood how to get the -­‐.56m I gave you a point for extra credit. 7 For questions 19 and 21 consider the following velocity time graph. 19. What is the acceleration during segment D of the motion? A) 0 m/s2 B) 1 m/s2 C) -­‐2m/s2 D) 3 m/s2 The slope of a linear curve on a velocity time graph represents the acceleration—it is a graphical representation of the relationship, a =
Δv
. We can estimate the acceleration from t
the graph by determining the rise (representing the change in velocity) and dividing it by the run (representing the time interval). In this case, the time interval of segment is one second, and during that time the velocity changes from 0 m/s to 3 m/s—a change of 3 m. So, the acceleration is a=
Δv 3m / s
=
= 3m / s 2
t
1s
Answer D 20. During how many segments of the motion is the acceleration a positive non-­‐zero value? A) 0 segments B) 1 segment C) 2 segments D) 3 segments As the slope of the velocity time represents the acceleration, if the acceleration is a positive non-­‐
zero value, the slope of curve must be positive. In this case the slope is positive during segments B, and D. So, the acceleration has a positive non-­‐zero value during two segments of the motion. Answer C 21. What is the average acceleration during the first 10 seconds of the motion? A) 0 m/s2 B) 1 m/s2 C) 2 m/s2 D) 3 m/s2 I messed up on this question. In this case the given time interval is ten seconds, and during that time the velocity changes from -­‐2 m/s to 0 m/s—a change of 2 m/. So the acceleration is 8 a=
Δv 2m / s
=
= .2m / s 2 When I wrote this question for some reason my brain thought the t
10s
initial velocity was 0 m/s instead of -­‐2 m/s, in which case acceleration would have been a=
Δv 0m / s
=
= 0m / s 2 . Because of this, I removed this question, although if you showed me t
10s
evidence that you understood how to get the .2m/s2 I gave you a point for extra credit. (26 points total corrected to 50 points) Free Response—Remember: equation—substitution—calculation for full credit: 22. An F-­‐16 fighter jet is put in flight to investigate a suddenly-­‐appearing, unidentified stationary radar target 184 km to the east of the Air Force base. At the instant its wheels leave the ground, its velocity is 175 m/s. It then constantly accelerates to its cruising velocity 600 m/s over a displacement of 2325 m. a. What is the acceleration of the F-­‐16 from the time it becomes airborne to the time it reaches its cruising velocity Things you could do to help gather information: Create a motion diagram—creating a motion diagram in a systematic and methodical fashion helps to visualize the motion and helps to make certain that the information collection gathers all information, even information that is not immediately explicit in the problem description. If you do this you should use a pattern of data collection: What is x0 What is x What is v0 What is v What is a What is t For this problem, a useful motion diagram would be: With this diagram you get a thorough visual representation of a positive increasing velocity, and positive acceleration. (If v is constant, just collect v) (Use arrows to illustrate the direction and change—if any—of velocity (Use an arrow to illustrate the direction of the accelerations—it will always be in the direction of increasing lengths of the velocity vectors (Don’t include on the motion diagram, just put in the given list) Create a given/need/formula/solve list—while the motion diagram is a good way to visualize the motion and make sure all data is collected, a given list organizes this information in a way that is useful in determining what formula(s) you may need to solve the problem. You could either (a) use the motion diagram to help you populate the given list, or (b) simply create the given list as you are reading through the problem. I would make it a habit that if you create the given list while reading through the problem, you try to isolate information in a particular order (x0, x, vo, v, a, t) that you always use. That way, if certain pieces of information are not obvious to you from 9 reading through the problem, you still make an effort to identify that information. If you just write down values as they come to you, you will forget to collect important information. A given list for this problem would look like: After collecting the given and needed information, it is easy to see that in order to solve for acceleration, as you don’t have time, you will need to use the velocity with constant acceleration/no time formula. Always isolate the desired variable before plugging in values. Then plug in the values and solve. There is only 1 SF in the answer. Information in the box is what I need to see for your answer. Answer 70 m/s2 (Note, as this is a multipart problem, I would show answers both to more than the correct number of significant figures and to the correct number—it is likely that you will need your answer in a subsequent part and you don’t want to introduce rounding error so you will probably need to use the answer with a greater number of SF’s.) Collect information using “internal dialogue” and use this to determine the formula needed. In the beginning it is most useful to practice collecting information by writing it down in a motion diagram and/or a given list of some type. However, you eventually want to get to the point where you can create a running list of information in your head and use this to select the formula you need without needing to write a lot of information down. In this case, as you read through the problem the first piece of information you come to is a displacement of 184 km, and you might start by thinking this is going to be x. But then you are given an initial and final velocity and a different displacement and asked to find acceleration. At this point in the internal dialogue you ask—do I have, or do I need time. If the answer is no, you know you need to use the velocity with constant acceleration formula, so write it down and isolate the desired variable, in this case a. (If you did have time, or needed time, the next question you would ask is, do I have or need final velocity—if the answer to this is no, you are committed to the displacement with constant acceleration/initial velocity formula). v 2 = v02 + 2a(x − x0 )
a=
v 2 − v02 (600m / s)2 − (175m / s)2
=
== 70.8m / s 2 = 70m / s 2 2x
2(2325m)
(3 points) b. How long did the F-­‐16 take to get to cruising velocity? If you are collecting data using a motion diagram and/or given list, all you would need to do is add t to the “Need” part of the list and one of the following formulas and solutions. At this point, you would be able to find time in one of two ways. As you already have an initial and final velocity, and an initial and final position, you could use the displacement with constant velocity formula and isolate time. 10 x = x0 + 12 (v + v0 )t
t=
2(x − x0 )
(v + v0 )
=
(2)(2325m)
= 6s (600m / s + 175m / s)
Or, as you have recently obtained acceleration, you could use the velocity with constant acceleration formula, isolating for time, as you also have initial and final velocity. If you use this approach you should use the unrounded value for acceleration (70.83 m/s2) rather than your final answer (70 m/s2) in order to avoid rounding error. The advantage to solving using the first approach is that it avoids using the calculated acceleration value all together, and so does not have the possibility of rounding error. In either case the answer should have 1 SF. v = v0 + at
t=
v − v0 600m / s − 175m / s
=
= 6s a
70.83m / s 2
Answer 6s (3 points) c. When the F-­‐16 reaches cruising velocity the unidentified object is still 183 km away. Assuming the object remains stationary, if the first visual siting of the object occurs when the F-­‐16 is still 17.0 km from the object, how long does it take for the pilot to get to the point where he first visualizes it, once he reaches cruising velocity (express the answer in minutes). If you used a motion diagram to gather information the thought process should be something like this. You are free to set the initial position as x0 = 0 m. For x, you are told that the UFO is 183 km away, but that the pilot can see it 17.0 km before he reaches it—therefore, x = 183 km – 17.0 km = 166 km = 166,000 m. (You should take care of changing this to m because you are eventually going to need to any way). Put these positions on the diagram. You also know that for this motion, the velocity has already been given to you—you were given that its cruising velocity is 600 m/s, and that it is a constant velocity. To put a constant velocity on the motion diagram you just use a single velocity arrow, rather than a series of increasing length arrows. Finally, as you are not accelerating you could state that a = 0 m/s2 on the diagram. This would be represented by a dot rather than an arrow. The diagram should look like this. With this diagram you get a thorough visual representation of a constant velocity of 600 m/s over a displacement of 166,000 m. If you created a given list, whether or not you created a motion diagram, it should look like this: From the given information it is easy to see that as you have a constant velocity, an initial and final position, and you need time, you should use the displacement with constant velocity formula, and isolate for time. Then, plug the values in and solve. 11 If you went through this process mentally, the first piece of information you come to is “still 183 km away.” Don’t automatically assume this is the displacement. You are then told that the real displacement is a point 17.0 km before you get to the UFO, so in your mind you are now thinking that the real “x” is 183 – 17=166 km=166,000m. You then know that you have an x0 (x0=0m) and an x. You are then told that the jet has been traveling at cruising velocity the whole time so there is no initial and final velocity, just a constant velocity of 600 m/s. You are then asked for time to travel the displacement of 166,000 m at the constant velocity of 600 m/s. Knowing that you have a displacement and a constant velocity you know you need to use our only “constant velocity” formula, the displacement with constant velocity formula, x = x0 + vt. Answer 5 min (3 points) x = x0 + vt
t=
x − x0
v
=
166,000m
1 min
= 277s x
= 4.6 min = 5 min 600 m / s
60 s
23. You are driving home on a weekend from school at 65 mi/h for 125 miles. It then starts to snow and you slow
to 30 mi/h. You arrive home after driving 6 hours and 20 minutes. How far is your hometown from school?
You read through the problem and find out that there are two different motions that you need to determine in order to determine the overall motion. Specifically, you want the overall displacement and you need to determine what the displacement is for each part of the motion and add these together. Working backwards: This is a good problem to work backward on. You know you need displacement—it should make sense that the overall displacement will be Δx =
Δx1 + Δx2
You already have
Δx1 = 125 mi
But, you don’t have
Δx2
But, you do know a formula that could given you this—as you are traveling at a
constant velocity during the second part of the motion, the displacement during
that time is
Δx2 = v2t2
You have velocity (30 mi/h) but you don’t have time
But, you have been given the overall time and you can find the time
of the first part of the motion and subtract it from the overall time—that
is t2 = t – t1
You do have the overall time (6.33 hr) but you don’t have t1
Because you know that the displacement for the first part of the motion
was 125 mi, and that the velocity during this time was 65 mi/h, you can
find t1 by rearranging the velocity formula.
Δx
Δx
v1 = 1 → t1 = 1
t1
v1
Then, plug the values in and solve for the time of the first part of the motion.
Then use this value to solve for the time of the second part of the motion.
Then use this value to solve for the displacement of the second part of the
motion. Then use this value to solve for the overall displacement.
The overall work should look something like this:
12 Δx = Δx1 + Δx2 = 125 mi + 132.3 mi = 257mi
Δx2 = v2t2 = (30mi / h)(4.41h) = 132.3 mi
t2 = t − t1 = 6.33 h − 1.92 h = 4.41 h
t1 =
Δx
125mi
=
= 1.92 h
v 65 mi / hr
Working forward through the problem: you could also work forward through the problem, but the work
you show is not likely to be nearly as “clean” and logical as if you worked backward.
As you read the problem whether you work forward or backward you realize that you need to find the
overall displacement, that there are two parts to the motion, and that you have been given the
displacement of the first part but need to find the displacement of the second part.
You see that you have been given a velocity of 30 mph for the second part of the motion and
know that in order to find the displacement you will need the time of the second part of the
motion, which you don’t have.
But, you do see that you have been given the time for the overall motion, and maybe
that you can find the time of the first part of the motion and subtract it from this, so
you will need to start by trying to find the time of the first part of the motion.
Because you know that
v1 =
Δx1
Δx
→ t1 = 1
t1
v1
you can plug the values in for the displacement and velocity to find the time. Then you can take the time for the first part of the motion and use it to find the time of the second part of the motion. Then you can take the time of the second part of the motion and use it to find displacement for the second part of the motion, which you can then use to find the overall displacement. The work for this should look something like: Δx
125mi
=
= 1.92 h
v 65 mi / hr
t2 = t − t1 = 6.33 h − 1.92 h = 4.41 h
t1 =
Δx2 = v2t2 = (30mi / h)(4.41h) = 132.3 mi
Δx = Δx1 + Δx2 = 125 mi + 132.3 mi = 257mi
basically the reverse of what the process was working backward. (4 points) 13 24. You are driving down a long, lonely, straight stretch of Interstate 70 late at night at a velocity of 40.0 m/s. You can see nothing except what is in front of you. Unknown to you (because you are grooving to She-­‐Daisy’s Greatest Hits in an attempt to stay awake), a Utah Highway Patrol officer begins to accelerate at a constant rate of 10.0 m/s2 in the same direction you are going, exactly at the same time you pass his vehicle. You do not detect his presence until he is right beside you, at which point he turns on his siren and lights. (a) This is one of those problems where we have two objects in motion and we are to compare
the motion of one object with the other. It is likely in these types of problems 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 police officer will be x for the car.
We have been given information about the motion of both vehicles so we should be
able to create formulas that would calculate the displacement for each vehicle at the
time the officer catches up. Think—what would writing such a displacement formula
for each vehicle involve.
For the officer, 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 car, 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 officer catches the car.
Then, because xcar would equal xoff, we could set their equations equal to each other
and solve for t. The result should have 3 sig figs. Answer 8.00 s (6 points)
Officer
xoff = x0 + v0t + 12 aoff t
1
2
1
2
xcar = x0 + vcar t
aoff t 2 = vcar t
aoff t = vcar
t=
Car
2
2vcar 2(40.0m / s)
=
= 8.00s
aoff
10.0m / s 2
14 (b) 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 officer—plug these in and solve. The result should have 3 sig figs.
Answer 80.0m/s; 179mph 4 points
v = v0 + at = (10.0m / s 2 )(8.00s) = 80.0m / s x
1km
.621m 3600s
x
x
= 179 mph
1000m 1km
1 hr
(c) 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 320 m 4 points
Officer xoff = x0 + v0t + 12 aoff t 2 = 12 (10.0m / s 2 )(8.00s)2 = 320m
Car
xcar = x0 + vcar t = (40.0m / s)(8.00s) = 320m
Total Free Response 27 points 15