Southern New Hampshire University
PHY-150 - Physics I with Lab
PHY-150 Module Two Homework | Questions & Complete
Solutions
Kinematics, Vectors, Projectile Motion, Circular Motion | SNHU | Title 64 | 2026 Update
Assessment Module Two Homework - WebAssign Coursework (Title 64)
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 Module Two Competencies, WebAssign Algorithmic Precision
CONFIDENTIALITY NOTICE: This document is prepared for academic reference purposes only. All questions, solutions, and
explanations are derived from SNHU PHY-150 Module Two curriculum and WebAssign coursework. Unauthorized distribution is
prohibited.
, SNHU PHY-150 | Module Two Homework | Title 64 | 2026
Section 1: Vectors, 1D Kinematics, and Graphical Analysis
Q1. A displacement vector has components Ax = 35.0 m and Ay = -25.0 m. What is the magnitude and direction (angle
from +x axis) of this vector?
A. 43.0 m, 35.5 degrees below +x [wrong direction: angle = arctan(25/35) = 35.5 but wrong sign handling]
B. 43.0 m, 35.5 degrees below +x axis (i.e., -35.5 degrees or 324.5 degrees) [CORRECT]
C. 60.0 m, 45.0 degrees below +x [wrong magnitude: used |A| = 35+25 = 60 m (magnitudes do not add this way)]
D. 35.0 m, 25.0 degrees below +x [wrong: used only the x-component as magnitude]
Correct Answer: B
Rationale: |A| = sqrt(A_x^2 + A_y^2) = sqrt(35^2 + 25^2) = sqrt(1225 + 625) = sqrt(1850) = 43.0 m. theta =
arctan(A_y/A_x) = arctan(-25/35) = -35.5 degrees (below +x axis, or 324.5 degrees measured counterclockwise). C adds
components directly instead of using the Pythagorean theorem. D uses only one component.
Q2. A car accelerates uniformly from rest at 3.20 m/s^2 for 8.00 s. How far does it travel?
A. 51.2 m [wrong - used d = v_0*t + a*t = 0 + 3.2*8 = 25.6, missing 1/2 factor and forgot to square t]
B. 25.6 m [wrong - used d = v_0*t + (1/2)*a*t = 0 + 0.5*3.2*8 = 12.8, forgot to square t]
C. 204.8 m [wrong - used d = v_0*t + a*t^2 = 0 + 3.2*64 = 204.8, forgot the 1/2 factor]
D. 102.4 m [CORRECT]
Correct Answer: D
Rationale: d = 0 + (1/2)(3.20)(8.00)^2 = (1/2)(3.20)(64.0) = 102.4 m. A misses both the 1/2 and t^2 factors. B misses t^2.
C misses the 1/2 factor.
Q3. A ball is thrown straight upward with an initial velocity of 25.0 m/s. How long does it take to reach maximum
height? (g = 9.80 m/s^2)
A. 2.55 s [CORRECT]
B. 5.10 s [wrong - used t = 2*v_0/g = 2*25/9.80 = 5.10 s, which is total time of flight, not time to max height]
C. 1.28 s [wrong - used t = v_0/(2g) = 25/(2*9.80) = 1.28 s, halved incorrectly]
D. 3.50 s [wrong - used approximate g = 10 and made arithmetic error]
Correct Answer: A
Rationale: At maximum height v = 0: v = v_0 - gt, so 0 = 25.0 - 9.80t, t = 25.0/9.80 = 2.551 s. B gives the total round-trip
time. C halves the formula incorrectly. D is a calculation error.
Q4. A velocity-time graph shows a straight line passing through (0, 20 m/s) and (4 s, 12 m/s) on the v-t axes. What is the
displacement from t = 0 to t = 4 s?
A. 32.0 m [wrong - used d = (v_i + v_f)/2 = 16, forgot to multiply by time]
B. 80.0 m [wrong - used d = v_i * t = 20*4 = 80 m, assuming constant velocity]
C. 64.0 m [CORRECT]
D. 128.0 m [wrong - used d = (v_i + v_f) * t = 32*4 = 128, forgot to divide by 2]
Correct Answer: C
Rationale: For constant acceleration, displacement equals the area under the v-t curve (trapezoid area): d = ((v_i + v_f)/2)
* delta_t = ((20 + 12)/2) * 4 = 16 * 4 = 64.0 m. A forgot to multiply by time. B assumed constant initial velocity. D forgot
to divide by 2.
Q5. An object moving in 1D has the position function x(t) = 3.0t^2 - 2.0t + 5.0 (x in meters, t in seconds). What is the
object's velocity at t = 3.0 s?
A. 52.0 m/s [wrong - used incorrect differentiation or evaluated a wrong expression]
B. 16.0 m/s [CORRECT]
C. 26.0 m/s [wrong - evaluated position x(3) = 3(9) - 2(3) + 5 = 26.0 m, confused position with velocity]
D. 34.0 m/s [wrong - used incorrect derivative formula]
Correct Answer: B
Rationale: v(t) = dx/dt = d/dt(3t^2 - 2t + 5) = 6t - 2. At t = 3: v = 6(3) - 2 = 18.0 - 2.0 = 16.0 m/s. C evaluates position
instead of taking the derivative. A and D use incorrect differentiation.
Q6. Vector A = 12.0 m at 30.0 degrees above +x axis. Vector B = 8.0 m at 60.0 degrees below +x axis. What is the
x-component of the resultant vector R = A + B?
A. 4.00 m [wrong - used only B_x, forgot A_x]
Page 2
PHY-150 - Physics I with Lab
PHY-150 Module Two Homework | Questions & Complete
Solutions
Kinematics, Vectors, Projectile Motion, Circular Motion | SNHU | Title 64 | 2026 Update
Assessment Module Two Homework - WebAssign Coursework (Title 64)
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 Module Two Competencies, WebAssign Algorithmic Precision
CONFIDENTIALITY NOTICE: This document is prepared for academic reference purposes only. All questions, solutions, and
explanations are derived from SNHU PHY-150 Module Two curriculum and WebAssign coursework. Unauthorized distribution is
prohibited.
, SNHU PHY-150 | Module Two Homework | Title 64 | 2026
Section 1: Vectors, 1D Kinematics, and Graphical Analysis
Q1. A displacement vector has components Ax = 35.0 m and Ay = -25.0 m. What is the magnitude and direction (angle
from +x axis) of this vector?
A. 43.0 m, 35.5 degrees below +x [wrong direction: angle = arctan(25/35) = 35.5 but wrong sign handling]
B. 43.0 m, 35.5 degrees below +x axis (i.e., -35.5 degrees or 324.5 degrees) [CORRECT]
C. 60.0 m, 45.0 degrees below +x [wrong magnitude: used |A| = 35+25 = 60 m (magnitudes do not add this way)]
D. 35.0 m, 25.0 degrees below +x [wrong: used only the x-component as magnitude]
Correct Answer: B
Rationale: |A| = sqrt(A_x^2 + A_y^2) = sqrt(35^2 + 25^2) = sqrt(1225 + 625) = sqrt(1850) = 43.0 m. theta =
arctan(A_y/A_x) = arctan(-25/35) = -35.5 degrees (below +x axis, or 324.5 degrees measured counterclockwise). C adds
components directly instead of using the Pythagorean theorem. D uses only one component.
Q2. A car accelerates uniformly from rest at 3.20 m/s^2 for 8.00 s. How far does it travel?
A. 51.2 m [wrong - used d = v_0*t + a*t = 0 + 3.2*8 = 25.6, missing 1/2 factor and forgot to square t]
B. 25.6 m [wrong - used d = v_0*t + (1/2)*a*t = 0 + 0.5*3.2*8 = 12.8, forgot to square t]
C. 204.8 m [wrong - used d = v_0*t + a*t^2 = 0 + 3.2*64 = 204.8, forgot the 1/2 factor]
D. 102.4 m [CORRECT]
Correct Answer: D
Rationale: d = 0 + (1/2)(3.20)(8.00)^2 = (1/2)(3.20)(64.0) = 102.4 m. A misses both the 1/2 and t^2 factors. B misses t^2.
C misses the 1/2 factor.
Q3. A ball is thrown straight upward with an initial velocity of 25.0 m/s. How long does it take to reach maximum
height? (g = 9.80 m/s^2)
A. 2.55 s [CORRECT]
B. 5.10 s [wrong - used t = 2*v_0/g = 2*25/9.80 = 5.10 s, which is total time of flight, not time to max height]
C. 1.28 s [wrong - used t = v_0/(2g) = 25/(2*9.80) = 1.28 s, halved incorrectly]
D. 3.50 s [wrong - used approximate g = 10 and made arithmetic error]
Correct Answer: A
Rationale: At maximum height v = 0: v = v_0 - gt, so 0 = 25.0 - 9.80t, t = 25.0/9.80 = 2.551 s. B gives the total round-trip
time. C halves the formula incorrectly. D is a calculation error.
Q4. A velocity-time graph shows a straight line passing through (0, 20 m/s) and (4 s, 12 m/s) on the v-t axes. What is the
displacement from t = 0 to t = 4 s?
A. 32.0 m [wrong - used d = (v_i + v_f)/2 = 16, forgot to multiply by time]
B. 80.0 m [wrong - used d = v_i * t = 20*4 = 80 m, assuming constant velocity]
C. 64.0 m [CORRECT]
D. 128.0 m [wrong - used d = (v_i + v_f) * t = 32*4 = 128, forgot to divide by 2]
Correct Answer: C
Rationale: For constant acceleration, displacement equals the area under the v-t curve (trapezoid area): d = ((v_i + v_f)/2)
* delta_t = ((20 + 12)/2) * 4 = 16 * 4 = 64.0 m. A forgot to multiply by time. B assumed constant initial velocity. D forgot
to divide by 2.
Q5. An object moving in 1D has the position function x(t) = 3.0t^2 - 2.0t + 5.0 (x in meters, t in seconds). What is the
object's velocity at t = 3.0 s?
A. 52.0 m/s [wrong - used incorrect differentiation or evaluated a wrong expression]
B. 16.0 m/s [CORRECT]
C. 26.0 m/s [wrong - evaluated position x(3) = 3(9) - 2(3) + 5 = 26.0 m, confused position with velocity]
D. 34.0 m/s [wrong - used incorrect derivative formula]
Correct Answer: B
Rationale: v(t) = dx/dt = d/dt(3t^2 - 2t + 5) = 6t - 2. At t = 3: v = 6(3) - 2 = 18.0 - 2.0 = 16.0 m/s. C evaluates position
instead of taking the derivative. A and D use incorrect differentiation.
Q6. Vector A = 12.0 m at 30.0 degrees above +x axis. Vector B = 8.0 m at 60.0 degrees below +x axis. What is the
x-component of the resultant vector R = A + B?
A. 4.00 m [wrong - used only B_x, forgot A_x]
Page 2
