MCAT Physics - Problem Drill 06: Translational Motion
Question No. 1 of 10
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needed (3) Pick the answer (4) Go back to review the core concept tutorial as needed.
1. An object falls from rest from a high building. It falls for some time, t, and
covers a distance, d. If a similar object falls for time, 2t, how far will it fall?
Question #01
(A)
(B)
(C)
(D)
d/2
d
2d
4d
A. Incorrect!
A greater time falling would lead to at least somewhat more distance covered than
with time t.
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Choice
B. Incorrect!
A greater time falling would lead to at least somewhat more distance covered than
with time t.
C. Incorrect!
This answer may seem reasonable, but the distance covered depends on t squared,
not just t.
D. Correct!
The distance covered depends on t squared. Since, d=vit+at 2/2, when the time is
doubled, the distance covered is quadrupled. Therefore 4d is the correct answer.
Consider your kinematics formulas. Specifically d = vit+at2/2. Since the distance
depends on time squared. Doubling the time will quadruple the distance since 2 2 is
4.
(D) is the correct answer.
Solution
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Question No. 2 of 10
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needed (3) Pick the answer (4) Go back to review the core concept tutorial as needed.
2. A mile is approximately 1609m. Calculate the average speed in m/s of a quick
runner that covers one mile in exactly 4 minutes.
Question #02
(A) 6.7 m/s
(B) 26.8 m/s
(C) 402 m/s
(D) 1609 m/s
A. Correct!
You need to convert minutes to seconds, one mile into meters, then use v = d/t. v
= 1609m / 240sec = 6.7 m/s
B. Incorrect!
You forgot to use four minutes instead of one.
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Choice
C. Incorrect!
You forgot to change minutes into seconds.
D. Incorrect!
That’s way too fast, faster than the speed of sound. Consider the number of
seconds in four minutes.
You need to convert minutes to seconds, one mile into meters, then use v = d/t.
1 mile = 1609 m
4 minutes = 240 seconds
V = 1609 m/ 240 sec = 6.7 m/s
The correct answer is (A).
Solution
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Question No. 3 of 10
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needed (3) Pick the answer (4) Go back to review the core concept tutorial as needed.
3. A car travels on a slick highway at 25 m/s. When applying the brakes, it can
slow at -6 m/s/s. How much time is needed to bring it to a complete halt?
Question #03
(A)
(B)
(C)
(D)
1.6 sec
4.2 sec
9.8 sec
58.8 sec
A. Incorrect!
Consider the acceleration formula/definition. Acceleration equal change in velocity
per change in time.
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Choice
B. Correct!
Acceleration equal change in velocity per change in time. Rearranging gives time
equals change in velocity divided by acceleration.
The change in velocity is 25m/s and the acceleration is -6 m/s/s. Dividing these
quantities give 4.2 seconds.
C. Incorrect!
That’s the acceleration due to gravity. The car isn’t falling to earth so we won’t use
that one here.
D. Incorrect!
Consider the acceleration formula/definition. Acceleration equal change in velocity
per change in time.
Given:
vi = 25 m/s
acceleration = -6 m/s2
vf = 0 m/s (halted)
Find: time =?
Solution
a = Δv/Δt= vf-vi/Δt
t = (vf-vi)/a
=(0-25 m/s) / -6m/s2
= 4.2 s
The correct answer is (B).
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Question No. 4 of 10
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needed (3) Pick the answer (4) Go back to review the core concept tutorial as needed.
4. A confused and befuddled hiker walks 3km North, then 1km West, then 2km
South, then 3km West, and finally 2km North. After all this, they sit down and try
to use physics to compute their location. How far have they moved from their
original starting position?
Question #04
(A) 5 km
(B) 11 km
(C) 15 km
(D) 45 km
A. Correct!
Add similar components. Add up all the North/South movements, getting a
resultant for that direction. Add up all the East/West movements, getting a
resultant for that direction. Finally, combine those two resultants to get his final
distance from the starting point. Use the Pythagorean theorem to get the
magnitude of the resultant.
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Choice
B. Incorrect!
For distances that are opposite the assigned positive must be made negative. For
example, if North is assigned as positive, any South movements must be negative.
C. Incorrect!
Don’t simply add all the values, direction must be taken into account.
D. Incorrect!
Don’t forget to take the square root when you are finding the final resultant.
Adding all the North/South movements give +3km North.
Adding all the East/West movements gives -4km West.
Using these two as the sides of a right triangle, find the hypotenuse with the
Pythagorean theorem.
c 2  a2  b2
c  a2  b2
Solution
c  (3km)2  (4km)2
c  25km2  5.0 km
It would be typical to assign North as positive, and East as positive. However, if
that were reversed, as long as it was consistently used, the same answer would be
found.
Also, the problem doesn’t ask for a direction, but if it were to be found,
trigonometry could be used to find whatever angle was asked.
The correct answer is (A).
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Question No. 5 of 10
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needed (3) Pick the answer (4) Go back to review the core concept tutorial as needed.
5. An object is thrown downwards at 20 m/s from the top of the Empire State
building in New York. This building is 321m tall. Neglecting the effects of air
resistance and friction, find how fast will the object be moving when it gets to the
ground below.
Question #05
(A) +81.8 m/s
(B) +76.7 n/s
(C) +20m/s
(D) +9.8m/s
A. Correct!
Notice that the initial velocity and acceleration due to gravity are both in the same
direction. They must have the same sign. In this case, they are both made to be
+. Use vf2=vi2 + 2ad here. The initial velocity is 20 m/s, the distance is 321m,
and the acceleration from gravity is 9.8 m/s2.
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Choice
B. Incorrect!
You must account for the initial velocity of 20 m/s downward.
C. Incorrect!
The object will not fall at a constant velocity due to the acceleration from gravity.
Use a formula that incorporates that fact.
D. Incorrect!
9.8 is the magnitude of the acceleration from gravity. However, that isn’t the same
as the velocity of the object. Identify the information given, along with the
unknown, and look for a formula that incorporates all of those.
Given:
vi = 20 m/s this velocity will arbitrarily be made +.
a = 9.8 m/s2. Since we have made the down direction +, this too is +.
d= 321m
vf= ?
Use:
v f  v i  2ad
2
2


v f  20m/s  2 9.8m/s2 321m
2
Solution
2


v f  400m2 /s 2  6290m2 /s 2  6690m2 /s 2
2
v f  6690m2 /s 2  81.8m/s
The correct answer is (A).
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Question No. 6 of 10
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needed (3) Pick the answer (4) Go back to review the core concept tutorial as needed.
6. A motorcyclist throws a ball into the air ahead of him, always staying under the
ball in order to catch it. If the motorcyclist’s speed is 25 m/s, and the initial speed
of the ball is 50 m/s, how long is the time of flight?
Question #06
(A)
(B)
(C)
(D)
4.3
5.0
7.1
8.8
s
s
s
s
A. Incorrect.
This is roughly half of the correct value.
B. Incorrect.
The velocity of the ball is not 25 km/s.
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Choice
C. Incorrect.
The ball is not thrown up at a 45° angle.
D. Correct!
Splitting the movement of the ball into x and y components and calculate the time
going upwards. The time of flight is twice as much.
First, find both horizontal and vertical components of the ball motion.
Since the ball is always under the motorcycle, we know what the x-component of
the ball’s initial velocity is. Since we have the horizontal component, and the
hypotenuse of 50 m/s, we can use the Pythagorean theorem to find the vertical
component.
c 2 =a2 +b2
 50m/s  =  25m/s  +  Vv 
2
2
2
Vv =43.3m/s
Solution
Next, use the definition of acceleration. The change in vertical velocity of the ball
will be twice the amount we calculated since it goes up, and returns downwards at
an equal speed. If t(up) is the time to take going upwards, the initial speed of
vertical velocity is Vi = 43.3 m/s and the final speed vertically is Vf = 0, we can use
the acceleration formula, where ∆V (vertical) = Vf - Viwher, to calculate the time
taken to go upwards, t(up):
a=
Δv
t
t(up) =
Δv
0m/s-43.3m/s
=
= 4.4 s
a
-9.8m/s/s
Hence, t(total) = t(up) + t(down) = 2 x t(up) = 8.8 s
(D) is the correct answer.
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Question No. 7 of 10
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needed (3) Pick the answer (4) Go back to review the core concept tutorial as needed.
7. A balloon is at an altitude of 80m. It is moving upwards at a constant 10m/s
when a package is dropped. How long does it take for the package to reach the
ground?
Question #7
(A)
(B)
(C)
(D)
-3.1 sec
4.0 sec
5.2sec
8.0 sec
A. Incorrect!
This is a solution for the quadratic equation obtained with the given information,
but it is an unrealistic answer since it’s a negative time. This is not the actual
answer.
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Choice
B. Incorrect!
This would be the time if there was no initial velocity of the package. However
there is so that must be taken into account.
C. Correct!
Substitute the given values into d=vit+at2/2. The given distance and acceleration
can be taken as negative since they are both downward. The initial velocity is
positive. Solve the quadratic equation for t.
D. Incorrect!
This would be the time if the package dropped at a constant velocity of 10 m/s.
The acceleration due to gravity is known. The initial distance is taken as negative
since the package will be falling down. The initial velocity is also the initial velocity
of the package. We want to find the time it takes for the package to hit the
ground. Use:
d=vi t+at 2 /2
This will be a quadratic equation. Substitute values and notice the coefficients for
the quadratic formula. The units have been dropped to clarify the quadratic
relationship.
-80=10t-9.8t 2 /2
-4.9t 2 +10t+80=0
Solution
This is a quadratic equation in the form of ax2 + bx + c = 0, where a = -4.9; b =
10; c = 80.
Substituting into the quadratic formula give:
-b± b2 -4ac -10± 102 -4(-4.9)80
=
2a
2(-4.9)
Solving this expression gives two solutions -3.1 and 5.2. The negative value will be
disregarded. The realistic positive answer is t = 5.2 sec.
You can plug that value back in to double check the equality.
Correct answer is (C).
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Question No. 8 of 10
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needed (3) Pick the answer (4) Go back to review the core concept tutorial as needed.
8. A manned motorboat heads west at 4.0 km/h across a south flowing river which
flows at 3 .0 km/h. If the manned motorboat travels for 12 min, how far has it
gone?
Question #08
(A)
(B)
(C)
(D)
0.60 km
0.80 km
1.0 km
1.4 km
A. Incorrect.
The speed of the boat must be taken into account.
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Choice
B. Incorrect.
The speed of the river current must be taken into account.
C. Correct!
See below for the correct solution.
D. Incorrect.
Splitting the movement of the cannonball into components would be a good idea.
First, find the net velocity, letting the x-direction be east, and letting the y-direction
be north. Then convert the total time in minutes into hours. Then use the
equation distance= (velocity)(time).
v x =-4.0 km/h
v y =-3.0 km/h
Solution
v= v 2x +v 2y = (-4.0)2 +(-3.0)2 =5.0 km/h
1h
1
=
h
60 min 5.0
 1 
d=vΔt=  5.0 km/h  × 
h  = 1.00 km
 5.0 
Δt= 12 min  ×
The correct answer is (C).
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Question No. 9 of 10
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needed (3) Pick the answer (4) Go back to review the core concept tutorial as needed.
9. A ball is dropped. Which of the plots below represent the velocity as a
function of displacement?
(A)
Question #09
(B)
(C)
(D)
A. Correct!
v 2 =2ad, so v d . We are looking for a square root relationship, which is what
the graph represents.
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Choice
B. Incorrect.
The relationship between velocity and displacement is not linear.
C. Incorrect.
This graph assumes v  d .
2
D. Incorrect.
The velocity is not constant.
We are looking for a square root relationship, which is what the graph
represents. Recall that Vf2=Vi2+2ad. Since velocity depends upon the square
root of d, the graph would look like selection A.
Solution
The correct answer is (A).
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Question No. 10 of 10
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needed (3) Pick the answer (4) Go back to review the core concept tutorial as needed.
10. Suppose that the position of an object as a function of time is x(t)=t3 -t+1 , where t
is in seconds and x is in meters. What is the average velocity of the object
between t = 1.0 s and t = 2.0 s?
Question #05
(A)
(B)
(C)
(D)
1.0
5.8
6.0
6.5
m/s
m/s
m/s
m/s
A. Incorrect.
Finding the displacement over the time interval would be a good idea.
B. Incorrect.
You cannot just evaluate the derivative at 1.5 s.
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Choice
C. Correct!
Use the formula that average velocity is the displacement divided by the time
interval, as shown in the solution below.
D. Incorrect.
You cannot take the derivative at 1.0 s and 2.0 s and average.
Δx x(2)-x(1)
=
Δt
2-1
x(1)=1
v=
x(2)=7
v=6 m/s
Solution
Answer (C) is correct.
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