PHY-150 - Introductory Physics
Module Two Kinematics Lab Report | Questions & Answers
1D & 2D Kinematics, Data Analysis, Graphing, Error Calculation | Title 66 | 2026 Update
Assessment Module Two Kinematics Lab Report - Questions & Answers (Title 66)
Questions 25 Multiple Choice (A-D) with Complete Solutions
Date July 8, 2026
Academic Year 2026-2027
Cognitive Level 30% Recall | 50% Application | 20% Analysis
Standards SNHU PHY-150 Experimental Physics Competencies, 2026 Lab Standards
, PHY-150 | Module Two Kinematics Lab Report | Title 66 | 2026
Section 1: Principles of Kinematics & Experimental Setup (Q1–Q7)
Q1. A dynamics cart is released from rest at the top of an inclined track. It rolls down 2.00 m, bounces off a stopper at
the bottom, and rolls back up 0.50 m before momentarily stopping. What is the cart’s net displacement from its starting
position?
A. 2.50 m — the total path length traveled by the cart
B. 0.50 m — the distance the cart rolled back uphill after bouncing
C. 2.00 m — the distance the cart traveled down the incline
D. 1.50 m — the net change in position: 2.00 m down minus 0.50 m up = 1.50 m from start [CORRECT]
Correct Answer: D
Rationale: Displacement is the straight-line distance from start to end position, including direction. The cart moves 2.00
m down, then 0.50 m back up, giving a net displacement of 2.00 - 0.50 = 1.50 m down the incline. A is the total distance
(scalar), not displacement (vector). B is only the rebound distance. C ignores the rebound entirely.
Q2. In a kinematics experiment, a cart’s position is recorded at t = 0.00 s (x = 0.000 m) and at t = 2.00 s (x = 1.740 m).
What physical quantity does the ratio ∆x/∆t = 0.870 m/s represent?
A. The instantaneous velocity of the cart at t = 2.00 s
B. The average velocity of the cart over the 2.00 s time interval [CORRECT]
C. The instantaneous acceleration of the cart
D. The total displacement of the cart from its starting position
Correct Answer: B
Rationale: The ratio ∆x/∆t gives the average velocity over the interval, not the instantaneous velocity at any single point
(which requires ∆t → 0). A confuses average with instantaneous. C is acceleration (change in velocity over time), not
velocity. D is displacement (1.740 m), a scalar quantity without the time division.
Q3. A cart accelerates from rest down an inclined plane with constant acceleration. Which kinematic equation relates the
cart’s displacement to the elapsed time without requiring the final velocity?
A. v = v■ + at
B. v² = v■² + 2a∆x
C. ∆x = v■t + ½at² [CORRECT]
D. v_avg = (v■ + v_f) / 2
Correct Answer: C
Rationale: Equation C directly relates displacement (∆x) to time (t) and acceleration (a) without requiring the final
velocity (v_f). A requires v_f to solve for displacement. B also requires v_f. D gives average velocity but still requires v_f
to compute displacement.
Q4. A dynamics track is inclined at an angle of 5.0° above the horizontal. Neglecting friction, what is the theoretical
acceleration of a cart released from rest on this incline? (Use g = 9.80 m/s².)
A. 0.854 m/s² (a = g sin θ = 9.80 × sin 5.0° = 9.80 × 0.0872 = 0.854 m/s²) [CORRECT]
B. 9.80 m/s² (the full gravitational acceleration, ignoring the incline angle)
C. 1.71 m/s² (double the correct value, a common sign/convention error)
D. 0.0872 m/s² (computes sin 5.0° but forgets to multiply by g)
Correct Answer: A
Rationale: On a frictionless incline, only the component of gravity parallel to the surface accelerates the cart: a = g sin θ =
9.80 × 0.08716 = 0.854 m/s². B uses g directly without resolving the incline component. C doubles the result (possibly
confusing with 2× factor in ½at²). D computes sin 5.0° = 0.0872 but omits multiplication by g.
Q5. A student plots the average position of a cart (from 5 trials) versus time and observes a parabolic curve. To extract
the experimental acceleration, the student should:
A. Compute the slope of the secant line from the origin to the last data point to find the average velocity
B. Plot position versus t² and determine the slope of the best-fit line; acceleration = 2 × slope [CORRECT]
C. Calculate the area under the position-time curve to find the total distance traveled
D. Divide each position value by its corresponding t² (x/t²) and average the results, which directly gives
acceleration
Correct Answer: B
Page 2
Module Two Kinematics Lab Report | Questions & Answers
1D & 2D Kinematics, Data Analysis, Graphing, Error Calculation | Title 66 | 2026 Update
Assessment Module Two Kinematics Lab Report - Questions & Answers (Title 66)
Questions 25 Multiple Choice (A-D) with Complete Solutions
Date July 8, 2026
Academic Year 2026-2027
Cognitive Level 30% Recall | 50% Application | 20% Analysis
Standards SNHU PHY-150 Experimental Physics Competencies, 2026 Lab Standards
, PHY-150 | Module Two Kinematics Lab Report | Title 66 | 2026
Section 1: Principles of Kinematics & Experimental Setup (Q1–Q7)
Q1. A dynamics cart is released from rest at the top of an inclined track. It rolls down 2.00 m, bounces off a stopper at
the bottom, and rolls back up 0.50 m before momentarily stopping. What is the cart’s net displacement from its starting
position?
A. 2.50 m — the total path length traveled by the cart
B. 0.50 m — the distance the cart rolled back uphill after bouncing
C. 2.00 m — the distance the cart traveled down the incline
D. 1.50 m — the net change in position: 2.00 m down minus 0.50 m up = 1.50 m from start [CORRECT]
Correct Answer: D
Rationale: Displacement is the straight-line distance from start to end position, including direction. The cart moves 2.00
m down, then 0.50 m back up, giving a net displacement of 2.00 - 0.50 = 1.50 m down the incline. A is the total distance
(scalar), not displacement (vector). B is only the rebound distance. C ignores the rebound entirely.
Q2. In a kinematics experiment, a cart’s position is recorded at t = 0.00 s (x = 0.000 m) and at t = 2.00 s (x = 1.740 m).
What physical quantity does the ratio ∆x/∆t = 0.870 m/s represent?
A. The instantaneous velocity of the cart at t = 2.00 s
B. The average velocity of the cart over the 2.00 s time interval [CORRECT]
C. The instantaneous acceleration of the cart
D. The total displacement of the cart from its starting position
Correct Answer: B
Rationale: The ratio ∆x/∆t gives the average velocity over the interval, not the instantaneous velocity at any single point
(which requires ∆t → 0). A confuses average with instantaneous. C is acceleration (change in velocity over time), not
velocity. D is displacement (1.740 m), a scalar quantity without the time division.
Q3. A cart accelerates from rest down an inclined plane with constant acceleration. Which kinematic equation relates the
cart’s displacement to the elapsed time without requiring the final velocity?
A. v = v■ + at
B. v² = v■² + 2a∆x
C. ∆x = v■t + ½at² [CORRECT]
D. v_avg = (v■ + v_f) / 2
Correct Answer: C
Rationale: Equation C directly relates displacement (∆x) to time (t) and acceleration (a) without requiring the final
velocity (v_f). A requires v_f to solve for displacement. B also requires v_f. D gives average velocity but still requires v_f
to compute displacement.
Q4. A dynamics track is inclined at an angle of 5.0° above the horizontal. Neglecting friction, what is the theoretical
acceleration of a cart released from rest on this incline? (Use g = 9.80 m/s².)
A. 0.854 m/s² (a = g sin θ = 9.80 × sin 5.0° = 9.80 × 0.0872 = 0.854 m/s²) [CORRECT]
B. 9.80 m/s² (the full gravitational acceleration, ignoring the incline angle)
C. 1.71 m/s² (double the correct value, a common sign/convention error)
D. 0.0872 m/s² (computes sin 5.0° but forgets to multiply by g)
Correct Answer: A
Rationale: On a frictionless incline, only the component of gravity parallel to the surface accelerates the cart: a = g sin θ =
9.80 × 0.08716 = 0.854 m/s². B uses g directly without resolving the incline component. C doubles the result (possibly
confusing with 2× factor in ½at²). D computes sin 5.0° = 0.0872 but omits multiplication by g.
Q5. A student plots the average position of a cart (from 5 trials) versus time and observes a parabolic curve. To extract
the experimental acceleration, the student should:
A. Compute the slope of the secant line from the origin to the last data point to find the average velocity
B. Plot position versus t² and determine the slope of the best-fit line; acceleration = 2 × slope [CORRECT]
C. Calculate the area under the position-time curve to find the total distance traveled
D. Divide each position value by its corresponding t² (x/t²) and average the results, which directly gives
acceleration
Correct Answer: B
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