PHYS 165 MODULE 3 EXAM — 1D
KINEMATICS, MOTION IN A
STRAIGHT LINE & FREE FALL
OFFICIAL PRACTICE EXAM 2026/2027
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SECTION 1: POSITION, DISPLACEMENT & VELOCITY Q1 – Q10
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Question 1 of 50
A student drives 4.0 km east to a library, then 2.5 km west to a coffee shop, and finally 1.5 km
east to the university campus. Taking east as the positive direction, what is the student's
displacement from home?
A. 8.0 km east
B. 3.0 km east ✓ CORRECT
C. 3.0 km west
D. 8.0 km west
Correct Answer: B
Rationale: Displacement is the net change in position from the starting point, found by adding
the eastward segments and subtracting the westward segment to yield 3.0 km east. Choice A
incorrectly adds all distances traveled without regard to direction, giving total distance rather
than displacement. Remember that displacement is a vector quantity that depends only on
initial and final positions, not the path taken.
Question 2 of 50
A test cart moves along a straight track, traveling 36 m in the first 6.0 s, then remaining at
rest for 4.0 s, then covering another 24 m in the next 8.0 s. What is the average velocity for
the entire trip?
A. 6.0 m/s
B. 5.0 m/s
C. 4.0 m/s
D. 3.3 m/s ✓ CORRECT
Correct Answer: D
,Rationale: Average velocity equals total displacement divided by total elapsed time, so 60 m
divided by 18 s gives 3.3 m/s. Choice B incorrectly divides displacement by the moving time
only, omitting the 4.0 s rest period during which position remained constant. Always use the
full interval from start to finish when computing average velocity.
Question 3 of 50
A remote-controlled car moves along a straight track, traveling 12 m north in 3.0 s, then
immediately reverses and travels 8.0 m south in 2.0 s. What is the average speed of the car
for the entire 5.0 s interval?
A. 4.0 m/s ✓ CORRECT
B. 0.80 m/s
C. 2.4 m/s
D. 6.0 m/s
Correct Answer: A
Rationale: Average speed is the total distance traveled divided by the elapsed time, so 20 m
divided by 5.0 s equals 4.0 m/s. Choice B gives the magnitude of average velocity, using net
displacement rather than total path length, which is the defining distinction between speed
and velocity. Speed is always a positive scalar that accumulates every segment of the
journey regardless of direction.
Question 4 of 50
The position of a particle moving along the x-axis is given by x(t) = 2.0t² - 5.0t + 3.0, where x is
in meters and t is in seconds. What is the instantaneous velocity at t = 3.0 s?
A. 1.0 m/s
B. 6.0 m/s
C. 7.0 m/s ✓ CORRECT
D. 12 m/s
Correct Answer: C
Rationale: Instantaneous velocity is the derivative of position with respect to time, giving v(t)
= 4.0t - 5.0, which evaluates to 7.0 m/s at t = 3.0 s. Choice B confuses instantaneous velocity
with the position coordinate at that instant, which is 6.0 m. For polynomial position
functions, differentiate term by term and substitute the specific time value.
Question 5 of 50
A jogger runs 400 m around a circular track and returns to the starting line in 80 s. Which
statement is true?
A. The jogger's distance is zero and displacement is 400 m
, B. The jogger's distance is 400 m and displacement is zero ✓ CORRECT
C. The jogger's average velocity is 5.0 m/s
D. The jogger's instantaneous velocity is always 5.0 m/s
Correct Answer: B
Rationale: After one complete lap, the jogger returns to the starting point, so displacement is
zero while the distance traveled equals the 400 m path length. Choice A reverses the
definitions of distance and displacement, a common error when working with closed-loop
paths. Displacement depends only on start and end positions, making it zero for any
complete closed path.
Question 6 of 50
A train moves along a straight section of track. It passes a milepost at 12.0 m/s and, 15 s
later, passes the next milepost 240 m away. What is the average velocity during this interval?
A. 16.0 m/s ✓ CORRECT
B. 14.0 m/s
C. 12.0 m/s
D. 20.0 m/s
Correct Answer: A
Rationale: Average velocity is defined as displacement divided by the time interval, so 240 m
divided by 15 s yields 16.0 m/s. Choice B incorrectly averages the initial and final velocities
without knowing the final speed or assuming constant acceleration, which is not stated in the
problem. The average velocity formula requires only total displacement and total time, not
intermediate speeds.
Question 7 of 50
A physics student establishes a coordinate system with the positive x-direction pointing
north. A car moves south at 25 m/s for 10 s, then turns around and moves north at 15 m/s for
20 s. What is the average velocity for the entire trip?
A. -5.0 m/s
B. +5.0 m/s
C. -1.7 m/s
D. +1.7 m/s ✓ CORRECT
Correct Answer: D
Rationale: Taking north as positive, the southward displacement is -250 m and the northward
displacement is +300 m, giving a net displacement of +50 m over 30 s and an average
velocity of +1.7 m/s. Choice C makes the same calculation but retains a negative sign, likely
from treating the southward leg as positive or the northward leg as negative, violating the
KINEMATICS, MOTION IN A
STRAIGHT LINE & FREE FALL
OFFICIAL PRACTICE EXAM 2026/2027
══════════════════════════════════════
SECTION 1: POSITION, DISPLACEMENT & VELOCITY Q1 – Q10
══════════════════════════════════════
Question 1 of 50
A student drives 4.0 km east to a library, then 2.5 km west to a coffee shop, and finally 1.5 km
east to the university campus. Taking east as the positive direction, what is the student's
displacement from home?
A. 8.0 km east
B. 3.0 km east ✓ CORRECT
C. 3.0 km west
D. 8.0 km west
Correct Answer: B
Rationale: Displacement is the net change in position from the starting point, found by adding
the eastward segments and subtracting the westward segment to yield 3.0 km east. Choice A
incorrectly adds all distances traveled without regard to direction, giving total distance rather
than displacement. Remember that displacement is a vector quantity that depends only on
initial and final positions, not the path taken.
Question 2 of 50
A test cart moves along a straight track, traveling 36 m in the first 6.0 s, then remaining at
rest for 4.0 s, then covering another 24 m in the next 8.0 s. What is the average velocity for
the entire trip?
A. 6.0 m/s
B. 5.0 m/s
C. 4.0 m/s
D. 3.3 m/s ✓ CORRECT
Correct Answer: D
,Rationale: Average velocity equals total displacement divided by total elapsed time, so 60 m
divided by 18 s gives 3.3 m/s. Choice B incorrectly divides displacement by the moving time
only, omitting the 4.0 s rest period during which position remained constant. Always use the
full interval from start to finish when computing average velocity.
Question 3 of 50
A remote-controlled car moves along a straight track, traveling 12 m north in 3.0 s, then
immediately reverses and travels 8.0 m south in 2.0 s. What is the average speed of the car
for the entire 5.0 s interval?
A. 4.0 m/s ✓ CORRECT
B. 0.80 m/s
C. 2.4 m/s
D. 6.0 m/s
Correct Answer: A
Rationale: Average speed is the total distance traveled divided by the elapsed time, so 20 m
divided by 5.0 s equals 4.0 m/s. Choice B gives the magnitude of average velocity, using net
displacement rather than total path length, which is the defining distinction between speed
and velocity. Speed is always a positive scalar that accumulates every segment of the
journey regardless of direction.
Question 4 of 50
The position of a particle moving along the x-axis is given by x(t) = 2.0t² - 5.0t + 3.0, where x is
in meters and t is in seconds. What is the instantaneous velocity at t = 3.0 s?
A. 1.0 m/s
B. 6.0 m/s
C. 7.0 m/s ✓ CORRECT
D. 12 m/s
Correct Answer: C
Rationale: Instantaneous velocity is the derivative of position with respect to time, giving v(t)
= 4.0t - 5.0, which evaluates to 7.0 m/s at t = 3.0 s. Choice B confuses instantaneous velocity
with the position coordinate at that instant, which is 6.0 m. For polynomial position
functions, differentiate term by term and substitute the specific time value.
Question 5 of 50
A jogger runs 400 m around a circular track and returns to the starting line in 80 s. Which
statement is true?
A. The jogger's distance is zero and displacement is 400 m
, B. The jogger's distance is 400 m and displacement is zero ✓ CORRECT
C. The jogger's average velocity is 5.0 m/s
D. The jogger's instantaneous velocity is always 5.0 m/s
Correct Answer: B
Rationale: After one complete lap, the jogger returns to the starting point, so displacement is
zero while the distance traveled equals the 400 m path length. Choice A reverses the
definitions of distance and displacement, a common error when working with closed-loop
paths. Displacement depends only on start and end positions, making it zero for any
complete closed path.
Question 6 of 50
A train moves along a straight section of track. It passes a milepost at 12.0 m/s and, 15 s
later, passes the next milepost 240 m away. What is the average velocity during this interval?
A. 16.0 m/s ✓ CORRECT
B. 14.0 m/s
C. 12.0 m/s
D. 20.0 m/s
Correct Answer: A
Rationale: Average velocity is defined as displacement divided by the time interval, so 240 m
divided by 15 s yields 16.0 m/s. Choice B incorrectly averages the initial and final velocities
without knowing the final speed or assuming constant acceleration, which is not stated in the
problem. The average velocity formula requires only total displacement and total time, not
intermediate speeds.
Question 7 of 50
A physics student establishes a coordinate system with the positive x-direction pointing
north. A car moves south at 25 m/s for 10 s, then turns around and moves north at 15 m/s for
20 s. What is the average velocity for the entire trip?
A. -5.0 m/s
B. +5.0 m/s
C. -1.7 m/s
D. +1.7 m/s ✓ CORRECT
Correct Answer: D
Rationale: Taking north as positive, the southward displacement is -250 m and the northward
displacement is +300 m, giving a net displacement of +50 m over 30 s and an average
velocity of +1.7 m/s. Choice C makes the same calculation but retains a negative sign, likely
from treating the southward leg as positive or the northward leg as negative, violating the
