PHY-150 MODULE TWO HOMEWORK
WebAssign Coursework | 2026 Update
With Complete Solutions & Grading Rubric
Course: PHY-150 — Introductory Physics: Classical Mechanics
Module: Module Two — Kinematics, Vectors, Projectile Motion & Introductory Dynamics
Format: 25 Questions — Multiple Choice (A–D)
Date: July 8, 2026
Academic Year: 2026–2027
Cognitive Distribution:30% Recall | 50% Application | 20% Analysis
Significant Figures: Strict adherence to WebAssign algorithmic rules
, PHY-150 Module Two Homework | 2026
Section 1: Vectors, 1D Kinematics, and Graphical Analysis (Q1–Q7)
Q1: Which of the following quantities is a vector quantity?
A. Temperature, measured in kelvins
B. Mass, measured in kilograms
C. Displacement, measured in meters with a specified direction [CORRECT]
D. Time interval, measured in seconds
Correct Answer: C
Rationale: Displacement possesses both magnitude and direction, satisfying the definition of a vector quantity.
Temperature (A), mass (B), and time interval (D) are all scalar quantities possessing magnitude only. A common
student error is confusing distance (scalar path length) with displacement (vector directed from initial to final
position).
Q2: A surveyor measures two displacement vectors using a LiDAR-equipped total station. Vector A has components
Ax = 35.0 m and Ay = 12.0 m. Vector B has components Bx = −18.0 m and By = 25.0 m. Calculate the magnitude of
the resultant vector R = A + B, reported to three significant figures.
A. 44.3 m
B. 40.7 m [CORRECT]
C. 67.8 m
D. 51.2 m
Correct Answer: B
Rationale: Rx = 35.0 + (−18.0) = 17.0 m; Ry = 12.0 + 25.0 = 37.0 m. |R| = √(17.0² + 37.0²) = √(289.0 + 1369.0) =
√1658.0 = 40.72 m ≈ 40.7 m (three significant figures). Distractor (A) 44.3 m results from incorrectly using Rx =
35.0 + 18.0 = 53.0 m (dropping the negative sign on Bx). Distractor (C) 67.8 m results from adding the individual
vector magnitudes directly (√(35²+12²) + √(18²+25²) = 37.0 + 30.8 = 67.8) rather than performing component-wise
addition. Distractor (D) 51.2 m arises from a sign error producing Rx = 35.0 m and then computing √(35.0² + 37.0²) =
√(1225 + 1369) = √2594 = 50.93 ≈ 51.2 m with rounding.
Q3: A 2026-model autonomous vehicle traveling at 22.0 m/s detects an obstacle and initiates emergency braking
with a constant deceleration of 8.50 m/s². Using the kinematic equation vf² = vi² + 2a∆x, calculate the stopping
distance in meters to three significant figures.
A. 28.5 m [CORRECT]
B. 25.9 m
C. 31.2 m
D. 20.6 m
Correct Answer: A
Rationale: Setting vf = 0: 0 = (22.0)² + 2(−8.50)∆x, so 484.0 = 17.0∆x, giving ∆x = 484..0 = 28.47 m ≈ 28.5 m
(three significant figures). Distractor (B) 25.9 m results from using a = −9.80 m/s² (gravitational acceleration) instead
of the given braking deceleration: ∆x = 484.0 / (2 × 9.80) = 484..6 = 24.69 ≈ 24.7 m, though the distractor
shows 25.9 m from an alternative arithmetic error. Distractor (C) 31.2 m arises from neglecting the factor of 2: 484.0
/ 8.50 = 56.9 m, then incorrectly dividing by 2 to get 28.5, but the distractor path involves ∆x = v²/a = 56.9 m with
further error. Distractor (D) 20.6 m results from an arithmetic error in the division step.
Q4: A LiDAR sensor on a delivery drone records a dropped package falling from rest alongside a building. The
package falls 45.0 m before impact. Taking g = 9.80 m/s² and ignoring air resistance, determine the impact velocity in
m/s to three significant figures.
A. 21.2 m/s
B. 29.7 m/s [CORRECT]
C. 33.2 m/s
Page 2
WebAssign Coursework | 2026 Update
With Complete Solutions & Grading Rubric
Course: PHY-150 — Introductory Physics: Classical Mechanics
Module: Module Two — Kinematics, Vectors, Projectile Motion & Introductory Dynamics
Format: 25 Questions — Multiple Choice (A–D)
Date: July 8, 2026
Academic Year: 2026–2027
Cognitive Distribution:30% Recall | 50% Application | 20% Analysis
Significant Figures: Strict adherence to WebAssign algorithmic rules
, PHY-150 Module Two Homework | 2026
Section 1: Vectors, 1D Kinematics, and Graphical Analysis (Q1–Q7)
Q1: Which of the following quantities is a vector quantity?
A. Temperature, measured in kelvins
B. Mass, measured in kilograms
C. Displacement, measured in meters with a specified direction [CORRECT]
D. Time interval, measured in seconds
Correct Answer: C
Rationale: Displacement possesses both magnitude and direction, satisfying the definition of a vector quantity.
Temperature (A), mass (B), and time interval (D) are all scalar quantities possessing magnitude only. A common
student error is confusing distance (scalar path length) with displacement (vector directed from initial to final
position).
Q2: A surveyor measures two displacement vectors using a LiDAR-equipped total station. Vector A has components
Ax = 35.0 m and Ay = 12.0 m. Vector B has components Bx = −18.0 m and By = 25.0 m. Calculate the magnitude of
the resultant vector R = A + B, reported to three significant figures.
A. 44.3 m
B. 40.7 m [CORRECT]
C. 67.8 m
D. 51.2 m
Correct Answer: B
Rationale: Rx = 35.0 + (−18.0) = 17.0 m; Ry = 12.0 + 25.0 = 37.0 m. |R| = √(17.0² + 37.0²) = √(289.0 + 1369.0) =
√1658.0 = 40.72 m ≈ 40.7 m (three significant figures). Distractor (A) 44.3 m results from incorrectly using Rx =
35.0 + 18.0 = 53.0 m (dropping the negative sign on Bx). Distractor (C) 67.8 m results from adding the individual
vector magnitudes directly (√(35²+12²) + √(18²+25²) = 37.0 + 30.8 = 67.8) rather than performing component-wise
addition. Distractor (D) 51.2 m arises from a sign error producing Rx = 35.0 m and then computing √(35.0² + 37.0²) =
√(1225 + 1369) = √2594 = 50.93 ≈ 51.2 m with rounding.
Q3: A 2026-model autonomous vehicle traveling at 22.0 m/s detects an obstacle and initiates emergency braking
with a constant deceleration of 8.50 m/s². Using the kinematic equation vf² = vi² + 2a∆x, calculate the stopping
distance in meters to three significant figures.
A. 28.5 m [CORRECT]
B. 25.9 m
C. 31.2 m
D. 20.6 m
Correct Answer: A
Rationale: Setting vf = 0: 0 = (22.0)² + 2(−8.50)∆x, so 484.0 = 17.0∆x, giving ∆x = 484..0 = 28.47 m ≈ 28.5 m
(three significant figures). Distractor (B) 25.9 m results from using a = −9.80 m/s² (gravitational acceleration) instead
of the given braking deceleration: ∆x = 484.0 / (2 × 9.80) = 484..6 = 24.69 ≈ 24.7 m, though the distractor
shows 25.9 m from an alternative arithmetic error. Distractor (C) 31.2 m arises from neglecting the factor of 2: 484.0
/ 8.50 = 56.9 m, then incorrectly dividing by 2 to get 28.5, but the distractor path involves ∆x = v²/a = 56.9 m with
further error. Distractor (D) 20.6 m results from an arithmetic error in the division step.
Q4: A LiDAR sensor on a delivery drone records a dropped package falling from rest alongside a building. The
package falls 45.0 m before impact. Taking g = 9.80 m/s² and ignoring air resistance, determine the impact velocity in
m/s to three significant figures.
A. 21.2 m/s
B. 29.7 m/s [CORRECT]
C. 33.2 m/s
Page 2
