1. A car starts with velocity 0 m/s and accelerates at 2 m/s^2 for 5 s. The distance covered is:
- 12.5 m
- 26 m
- 50 m
- 25 m
Answer: D) 25 m
Explanation: s = ut + 12at^2 = 0(5) + 12(2)(5)^2 = 25 m. SLO P-11-B-04 · Application
MCQs / Physics / Kinematics and Projectile Motion
79 multiple-choice questions on Kinematics and Projectile Motion from Physics, each with the correct answer and, where available, a teacher's explanation. Read through them to revise the chapter, then practise the same questions one by one to test yourself.
Answer: D) 25 m
Explanation: s = ut + 12at^2 = 0(5) + 12(2)(5)^2 = 25 m. SLO P-11-B-04 · Application
Answer: A) its speed, shape and cross-sectional area
Explanation: Faster and larger objects experience more drag. SLO P-11-B-06 · Understanding
Answer: B) an overestimate of range and height
Explanation: Real drag always reduces them. SLO P-11-B-06 · Understanding
Answer: D) air resistance equals its weight (terminal velocity)
Explanation: The net force becomes zero. SLO P-11-B-06 · Understanding
Answer: C) decreases during the flight
Explanation: Drag opposes the horizontal motion. SLO P-11-B-06 · Understanding
Answer: B) less than g
Explanation: Drag now acts upward, against gravity. SLO P-11-B-06 · Understanding
Answer: D) 4.04 s
Explanation: h = 12gt^2 Rightarrow t = 2hg = 2(80)9.8 = 4.04 s. SLO P-11-B-04 · Application
Answer: D) 61.2 m
Explanation: R = u^2sin 2thetag = 61.2 m. SLO P-11-B-05 · Application
Answer: D) v^2 = u^2 + 2as
Explanation: It was obtained by eliminating t. SLO P-11-B-03 · Understanding
Answer: D) greater than g
Explanation: Both gravity and drag act downward. SLO P-11-B-06 · Understanding
Answer: B) 5 m/s^2
Explanation: 72 km/h = 20 m/s; a = 20/4 = 5 m/s^2. SLO P-11-B-04 · Application
Answer: A) lower than without air resistance
Explanation: Drag acts downward during the upward part, adding to gravity. SLO P-11-B-06 · Understanding
Answer: A) 17 m/s
Explanation: v = u + at = 5 + (1.5)(8) = 17 m/s. SLO P-11-B-04 · Application
Answer: C) a heavy, small, smooth metal ball
Explanation: High mass and small area give a small drag acceleration. SLO P-11-B-06 · Understanding
Answer: B) increasing velocity (acceleration)
Explanation: The slope (velocity) is increasing. SLO P-11-B-03 · Understanding
Answer: A) u + v2
Explanation: The velocity changes linearly, so the average is the mean of the end values. SLO P-11-B-03 · Understanding
Answer: C) 20.2 m
Explanation: Fall time t = 2h/g = 2.02 s; horizontal distance = vt = 20.2 m (horizontal velocity stays constant). SLO P-11-B-05 · Application
Answer: A) s = ut + 12at^2
Explanation: s = u + u + at2t = ut + 12at^2. SLO P-11-B-03 · Understanding
Answer: C) 10 m/s
Explanation: v = u + at = 0 + (2)(5) = 10 m/s. SLO P-11-B-04 · Application
Answer: D) 31.9 m
Explanation: H = u^2sin^2theta2g = 31.9 m. SLO P-11-B-05 · Application
Answer: D) air resistance and the release height favour a lower angle
Explanation: Both effects reduce the best angle below 45^circ. SLO P-11-B-06 · Understanding
Answer: D) 6 s
Explanation: Time up = u/g = 3 s; total time = 2u/g = 6 s. SLO P-11-B-04 · Application
Answer: C) 29.7 m/s
Explanation: v = 2gh = 2(9.8)(45) = 29.7 m/s. SLO P-11-B-04 · Application
Answer: C) 8 m/s
Explanation: v = u + at = 20 + (-2)(6) = 8 m/s. SLO P-11-B-04 · Application
Answer: C) a parabola
Explanation: y = xtantheta - gx^22u^2cos^2theta is quadratic in x. SLO P-11-B-05 · Application
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Practise Kinematics and Projectile Motion MCQs