AP Physics 1 Unit 1 Progress Check–style
Exam Questions and Answers with
Rationales Top Rated A+
AP Physics 1 — Unit 1 Practice Exam
Unit 1: Kinematics
Question 1
A car travels along a straight road. Its position as a function of time is shown by a graph in which
the slope of the position-time curve is positive and increasing.
Which statement best describes the car's motion?
A. The car is moving with constant positive velocity.
B. The car is moving with increasing positive velocity.
C. The car is moving with decreasing positive velocity.
D. The car is stationary.
Answer: B
Rationale:
The slope of a position-versus-time graph represents velocity. A positive slope means positive
velocity, while an increasing slope means the velocity is becoming larger. Therefore, the car has
increasing positive velocity, indicating positive acceleration.
Question 2
A runner starts from rest and accelerates uniformly at 2.0 m/s22.0\,\text{m/s}^2. What is the
runner's speed after 5.0 s?
A. 2.5 m/s
B. 5.0 m/s
C. 10 m/s
D. 25 m/s
,Answer: C
Rationale:
Use
v=v0+atv=v_0+at
Since v0=0v_0=0,
v=(2.0)(5.0)=10 m/sv=(2.0)(5.0)=10\,\text{m/s}
Thus, the correct answer is C.
Question 3
An object moves with constant velocity of 6 m/s6\,\text{m/s} for 8 s. How far does it travel?
A. 0.75 m
B. 14 m
C. 48 m
D. 64 m
Answer: C
Rationale:
For constant velocity,
Δx=vΔt\Delta x=v\Delta t Δx=(6)(8)=48 m\Delta x=(6)(8)=48\,\text{m}
Therefore, C is correct.
Question 4
A velocity-time graph for an object is a horizontal line at +4 m/s+4\,\text{m/s}. What does this
indicate?
A. The object has zero velocity.
B. The object has constant positive velocity.
C. The object has constant positive acceleration.
D. The object has increasing velocity.
Answer: B
,Rationale:
The vertical coordinate of a velocity-time graph represents velocity. A horizontal line means
velocity does not change, so acceleration is zero. Because the line is above zero, the velocity is
positive.
Question 5
An object has a velocity of 10 m/s10\,\text{m/s} at t=2 st=2\,\text{s} and 4 m/s4\,\text{m/s} at
t=5 st=5\,\text{s}. What is its average acceleration?
A. −2 m/s2-2\,\text{m/s}^2
B. −1.2 m/s2-1.2\,\text{m/s}^2
C. 1.2 m/s21.2\,\text{m/s}^2
D. 2 m/s22\,\text{m/s}^2
Answer: B
Rationale:
aavg=ΔvΔta_{\text{avg}}=\frac{\Delta v}{\Delta t}
aavg=4−105−2=−63=−2 m/s2a_{\text{avg}}=\frac{4-10}{5-2} =\frac{-6}{3} =-2\,\text{m/s}^2
So the correct answer is actually A.
Question 6
A ball is thrown vertically upward. At the highest point of its trajectory, which statement is
correct?
A. Its velocity and acceleration are both zero.
B. Its velocity is zero, but its acceleration is downward.
C. Its velocity is upward and its acceleration is zero.
D. Its velocity and acceleration are both upward.
Answer: B
Rationale:
At the highest point, the instantaneous velocity is zero because the ball changes direction.
However, gravity continues to act, producing a downward acceleration of approximately
9.8 m/s29.8\,\text{m/s}^2.
, Question 7
Two cars travel in the same direction. Car A has a velocity of 20 m/s20\,\text{m/s}, while Car B
has a velocity of 12 m/s12\,\text{m/s}. What is the velocity of Car A relative to Car B?
A. 8 m/s8\,\text{m/s}
B. 12 m/s12\,\text{m/s}
C. 20 m/s20\,\text{m/s}
D. 32 m/s32\,\text{m/s}
Answer: A
Rationale:
vA/B=vA−vBv_{A/B}=v_A-v_B vA/B=20−12=8 m/sv_{A/B}=20-12=8\,\text{m/s}
Thus, Car A moves at 8 m/s relative to Car B.
Question 8
An object starts from rest and travels 50 m50\,\text{m} in 5 s with constant acceleration. What is
its acceleration?
A. 2 m/s22\,\text{m/s}^2
B. 4 m/s24\,\text{m/s}^2
C. 5 m/s25\,\text{m/s}^2
D. 10 m/s210\,\text{m/s}^2
Answer: B
Rationale:
Use
Δx=v0t+12at2\Delta x=v_0t+\frac12at^2
Because the object starts from rest,
50=12a(5)250=\frac12a(5)^2 50=12.5a50=12.5a a=4 m/s2a=4\,\text{m/s}^2
Therefore, B is correct.
Exam Questions and Answers with
Rationales Top Rated A+
AP Physics 1 — Unit 1 Practice Exam
Unit 1: Kinematics
Question 1
A car travels along a straight road. Its position as a function of time is shown by a graph in which
the slope of the position-time curve is positive and increasing.
Which statement best describes the car's motion?
A. The car is moving with constant positive velocity.
B. The car is moving with increasing positive velocity.
C. The car is moving with decreasing positive velocity.
D. The car is stationary.
Answer: B
Rationale:
The slope of a position-versus-time graph represents velocity. A positive slope means positive
velocity, while an increasing slope means the velocity is becoming larger. Therefore, the car has
increasing positive velocity, indicating positive acceleration.
Question 2
A runner starts from rest and accelerates uniformly at 2.0 m/s22.0\,\text{m/s}^2. What is the
runner's speed after 5.0 s?
A. 2.5 m/s
B. 5.0 m/s
C. 10 m/s
D. 25 m/s
,Answer: C
Rationale:
Use
v=v0+atv=v_0+at
Since v0=0v_0=0,
v=(2.0)(5.0)=10 m/sv=(2.0)(5.0)=10\,\text{m/s}
Thus, the correct answer is C.
Question 3
An object moves with constant velocity of 6 m/s6\,\text{m/s} for 8 s. How far does it travel?
A. 0.75 m
B. 14 m
C. 48 m
D. 64 m
Answer: C
Rationale:
For constant velocity,
Δx=vΔt\Delta x=v\Delta t Δx=(6)(8)=48 m\Delta x=(6)(8)=48\,\text{m}
Therefore, C is correct.
Question 4
A velocity-time graph for an object is a horizontal line at +4 m/s+4\,\text{m/s}. What does this
indicate?
A. The object has zero velocity.
B. The object has constant positive velocity.
C. The object has constant positive acceleration.
D. The object has increasing velocity.
Answer: B
,Rationale:
The vertical coordinate of a velocity-time graph represents velocity. A horizontal line means
velocity does not change, so acceleration is zero. Because the line is above zero, the velocity is
positive.
Question 5
An object has a velocity of 10 m/s10\,\text{m/s} at t=2 st=2\,\text{s} and 4 m/s4\,\text{m/s} at
t=5 st=5\,\text{s}. What is its average acceleration?
A. −2 m/s2-2\,\text{m/s}^2
B. −1.2 m/s2-1.2\,\text{m/s}^2
C. 1.2 m/s21.2\,\text{m/s}^2
D. 2 m/s22\,\text{m/s}^2
Answer: B
Rationale:
aavg=ΔvΔta_{\text{avg}}=\frac{\Delta v}{\Delta t}
aavg=4−105−2=−63=−2 m/s2a_{\text{avg}}=\frac{4-10}{5-2} =\frac{-6}{3} =-2\,\text{m/s}^2
So the correct answer is actually A.
Question 6
A ball is thrown vertically upward. At the highest point of its trajectory, which statement is
correct?
A. Its velocity and acceleration are both zero.
B. Its velocity is zero, but its acceleration is downward.
C. Its velocity is upward and its acceleration is zero.
D. Its velocity and acceleration are both upward.
Answer: B
Rationale:
At the highest point, the instantaneous velocity is zero because the ball changes direction.
However, gravity continues to act, producing a downward acceleration of approximately
9.8 m/s29.8\,\text{m/s}^2.
, Question 7
Two cars travel in the same direction. Car A has a velocity of 20 m/s20\,\text{m/s}, while Car B
has a velocity of 12 m/s12\,\text{m/s}. What is the velocity of Car A relative to Car B?
A. 8 m/s8\,\text{m/s}
B. 12 m/s12\,\text{m/s}
C. 20 m/s20\,\text{m/s}
D. 32 m/s32\,\text{m/s}
Answer: A
Rationale:
vA/B=vA−vBv_{A/B}=v_A-v_B vA/B=20−12=8 m/sv_{A/B}=20-12=8\,\text{m/s}
Thus, Car A moves at 8 m/s relative to Car B.
Question 8
An object starts from rest and travels 50 m50\,\text{m} in 5 s with constant acceleration. What is
its acceleration?
A. 2 m/s22\,\text{m/s}^2
B. 4 m/s24\,\text{m/s}^2
C. 5 m/s25\,\text{m/s}^2
D. 10 m/s210\,\text{m/s}^2
Answer: B
Rationale:
Use
Δx=v0t+12at2\Delta x=v_0t+\frac12at^2
Because the object starts from rest,
50=12a(5)250=\frac12a(5)^2 50=12.5a50=12.5a a=4 m/s2a=4\,\text{m/s}^2
Therefore, B is correct.
