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Motion in 2D — NEET Physics MCQs with Solutions
Free NEET Physics Motion in 2D MCQs with step-by-step solutions covering Basics of 2D, Projectile Motion, Relative Motion in 2D, Rain Man Problem, River Boat Problem, Wind Problem. Practise online on Prepizo — no login needed.
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Sample questions with solutions
Q1 — Basics of 2D · easy · numerical
A particle moves from the point (1 m, 2 m) to the point (4 m, 6 m). The magnitude of its displacement is:
A. 3 m
B. 7 m
C. 4 m
D. 5 m ✓ Correct
Solution: Δx = 3 m and Δy = 4 m, so Δr = √(3² + 4²) = 5 m.
Q2 — Basics of 2D · easy · numerical
A bird flies from the point (5 m, -2 m) to the point (11 m, 6 m). The magnitude of its displacement is:
A. 8 m
B. 14 m
C. 6 m
D. 10 m ✓ Correct
Solution: Δx = 6 m and Δy = 8 m, so Δr = √(6² + 8²) = 10 m.
Q3 — Basics of 2D · easy · numerical
A person walks 30 m east and then 40 m north, taking a total of 10 s. The magnitude of the average velocity is:
A. 7 m/s
B. 4 m/s
C. 3 m/s
D. 5 m/s ✓ Correct
Solution: Displacement = √(30² + 40²) = 50 m; average velocity = 50/10 = 5 m/s.
Q4 — Basics of 2D · easy · numerical
At one instant a football has velocity components 9 m/s horizontally and 12 m/s vertically. Its speed at that instant is:
A. 3 m/s
B. 21 m/s
C. 12 m/s
D. 15 m/s ✓ Correct
Solution: Speed = √(9² + 12²) = √225 = 15 m/s.
Q5 — Basics of 2D · easy · numerical
The position of a particle is r = (4t î + 3t ĵ) m, where t is in seconds. The speed of the particle is:
A. 7 m/s
B. 1 m/s
C. 25 m/s
D. 5 m/s ✓ Correct
Solution: vₓ = 4 m/s and v_y = 3 m/s, so speed = √(4² + 3²) = 5 m/s.
Q6 — Basics of 2D · easy · numerical
The position of a particle is r = (5t î + 2t² ĵ) m, where t is in seconds. The magnitude of its acceleration is:
A. 9 m/s²
B. 4 m/s² ✓ Correct
C. 5 m/s²
D. 2 m/s²
Solution: aₓ = 0 and a_y = d²(2t²)/dt² = 4 m/s², so |a| = 4 m/s².
Q7 — Basics of 2D · easy · numerical
The velocity of a scooter changes from 10î m/s to 10ĵ m/s in 5 s. The magnitude of the average acceleration is:
A. 2√2 ≈ 2.8 m/s² ✓ Correct
B. 2 m/s²
C. 4 m/s²
D. 0 m/s²
Solution: Δv = 10ĵ − 10î, |Δv| = 10√2 m/s; average acceleration = 10√2/5 = 2√2 ≈ 2.8 m/s².
Q8 — Basics of 2D · easy · numerical
A particle starts from rest and moves with constant acceleration a = (3î + 4ĵ) m/s². Its speed after 2 s is:
A. 10 m/s ✓ Correct
B. 14 m/s
C. 7 m/s
D. 5 m/s
Solution: |a| = √(3² + 4²) = 5 m/s²; from rest, v = |a|t = 5 × 2 = 10 m/s.
Q9 — Basics of 2D · easy · numerical
The velocity of a particle is v = (6î + 6√3 ĵ) m/s. The angle its velocity makes with the x-axis is:
A. 60° ✓ Correct
B. 30°
C. 45°
D. 90°
Solution: tanθ = v_y/vₓ = 6√3/6 = √3, so θ = 60°.
Q10 — Basics of 2D · easy · numerical
A particle starts from the origin with x = 6t and y = 8t (both in metres, t in seconds). Its distance from the origin at t = 2 s is:
A. 14 m
B. 20 m ✓ Correct
C. 28 m
D. 10 m
Solution: At t = 2 s, x = 12 m and y = 16 m, so distance = √(12² + 16²) = 20 m.
Q11 — Basics of 2D · easy · numerical
The velocity of a cyclist makes 60° with the x-axis. If the x-component of the velocity is 10 m/s, the speed of the cyclist is:
A. 10√3 ≈ 17.3 m/s
B. 20 m/s ✓ Correct
C. 5 m/s
D. 40 m/s
Solution: v = vₓ/cos60° = 10/0.5 = 20 m/s.
Q12 — Basics of 2D · easy · numerical
A car starts at the origin and 2 s later is at a point 5 m east and 12 m north of it. The magnitude of its average velocity is:
A. 3.25 m/s
B. 13 m/s
C. 6.5 m/s ✓ Correct
D. 8.5 m/s
Solution: Displacement = √(5² + 12²) = 13 m; average velocity = 13/2 = 6.5 m/s.
Q13 — Projectile Motion · easy · numerical
A ball is projected with a speed of 20 m/s at 30° above the horizontal. Its time of flight is (g = 10 m/s²):
A. 1 s
B. 2 s ✓ Correct
C. 4 s
D. 2√3 ≈ 3.5 s
Solution: T = 2u sinθ/g = (2 × 20 × 0.5)/10 = 2 s.
Q14 — Projectile Motion · easy · numerical
A football is kicked with a speed of 40 m/s at 30° above the horizontal. The maximum height reached is (g = 10 m/s²):
A. 60 m
B. 40 m
C. 20 m ✓ Correct
D. 80 m
Solution: H = u²sin²θ/(2g) = (1600 × 0.25)/20 = 20 m.
Q15 — Projectile Motion · easy · numerical
A ball is projected at 20 m/s at an angle of 45° with the ground. Its horizontal range is (g = 10 m/s²):
A. 20 m
B. 10 m
C. 40 m ✓ Correct
D. 80 m
Solution: R = u²sin2θ/g = (400 × sin90°)/10 = 40 m.
Q16 — Projectile Motion · easy · numerical
A cricket ball can be thrown with a maximum speed of 30 m/s. The greatest horizontal distance it can cover is (g = 10 m/s²):
A. 45 m
B. 30 m
C. 90 m ✓ Correct
D. 180 m
Solution: Maximum range occurs at 45°: R_max = u²/g = 900/10 = 90 m.
Q17 — Projectile Motion · easy · numerical
A projectile is fired at 50 m/s at 60° above the horizontal. Its speed at the highest point is:
A. 25 m/s ✓ Correct
B. 50 m/s
C. 0 m/s
D. 25√3 ≈ 43.3 m/s
Solution: At the top only the horizontal component remains: v = u cos60° = 50 × 0.5 = 25 m/s.
Q18 — Projectile Motion · easy · numerical
A ball is thrown horizontally at 10 m/s from the top of a 45 m tower. The time it takes to reach the ground is (g = 10 m/s²):
A. 3 s ✓ Correct
B. 4.5 s
C. 2 s
D. 9 s
Solution: t = √(2h/g) = √(2 × 45/10) = √9 = 3 s (independent of the horizontal speed).
Q19 — Projectile Motion · easy · numerical
A stone is thrown horizontally at 12 m/s from the top of a 20 m building. The horizontal distance from the base at which it lands is (g = 10 m/s²):
A. 40 m
B. 24 m ✓ Correct
C. 12 m
D. 48 m
Solution: t = √(2 × 20/10) = 2 s, so R = 12 × 2 = 24 m.
Q20 — Projectile Motion · easy · numerical
A plane flying horizontally at 100 m/s at a height of 500 m releases a bomb. The bomb strikes the ground at a horizontal distance from the release point of (g = 10 m/s²):
A. 2000 m
B. 1000 m ✓ Correct
C. 500 m
D. 100 m
Solution: t = √(2 × 500/10) = 10 s; the bomb keeps the plane speed, so x = 100 × 10 = 1000 m.
Q21 — Projectile Motion · easy · numerical
A ball projected at 30° above the horizontal rises to a maximum height of 5 m. Its speed of projection is (g = 10 m/s²):
A. 10 m/s
B. 10√2 ≈ 14.1 m/s
C. 40 m/s
D. 20 m/s ✓ Correct
Solution: H = u²sin²30°/(2g): 5 = u² × 0.25/20, so u² = 400 and u = 20 m/s.
Q22 — Projectile Motion · easy · numerical
The horizontal range of a football is 4 times its maximum height. The angle of projection is:
A. 60°
B. 30°
C. 45° ✓ Correct
D. 75°
Solution: tanθ = 4H/R; with R = 4H, tanθ = 1, so θ = 45°.
Q23 — Projectile Motion · easy · numerical
Two balls are projected with the same speed, one at 30° and the other at 60° above the horizontal. The ratio of their maximum heights (30° to 60°) is:
A. 1 : 3 ✓ Correct
B. 1 : 2
C. 1 : √3
D. 3 : 1
Solution: H ∝ sin²θ, so the ratio is sin²30° : sin²60° = (1/4) : (3/4) = 1 : 3.
Q24 — Projectile Motion · easy · numerical
A ball thrown at 20 m/s at 30° above the horizontal has a certain range. With the same speed, the other angle of projection giving the same range is:
A. 90°
B. 60° ✓ Correct
C. 75°
D. 45°
Solution: Ranges are equal at complementary angles: 90° − 30° = 60°.
Q25 — Relative Motion in 2D · easy · numerical
Cyclist A rides due east at 6 m/s and cyclist B rides due north at 8 m/s. The magnitude of the velocity of A relative to B is:
A. 2 m/s
B. 7 m/s
C. 14 m/s
D. 10 m/s ✓ Correct
Solution: The velocities are perpendicular, so |v_AB| = √(6² + 8²) = 10 m/s.
Q26 — Relative Motion in 2D · easy · numerical
A train moves due east at 30 km/h while a car on a nearby road moves due north at 40 km/h. The speed of the train relative to the car is:
A. 50 km/h ✓ Correct
B. 10 km/h
C. 35 km/h
D. 70 km/h
Solution: Relative speed = √(30² + 40²) = √2500 = 50 km/h.
Q27 — Relative Motion in 2D · easy · numerical
A scooter moves due east at 12 m/s while a bird flies due north at 5 m/s. The speed of the scooter relative to the bird is:
A. 10 m/s
B. 7 m/s
C. 17 m/s
D. 13 m/s ✓ Correct
Solution: Relative speed = √(12² + 5²) = √169 = 13 m/s.
Q28 — Relative Motion in 2D · easy · numerical
Car A travels east at 20 m/s and car B travels west at 10 m/s on the same road. The speed of A relative to B is:
A. 30 m/s ✓ Correct
B. 15 m/s
C. 20 m/s
D. 10 m/s
Solution: For opposite directions the speeds add: v_AB = 20 + 10 = 30 m/s.
Q29 — Relative Motion in 2D · easy · numerical
Train A moving at 25 m/s follows train B moving at 15 m/s in the same direction on parallel tracks. The speed of A relative to B is:
A. 20 m/s
B. 25 m/s
C. 10 m/s ✓ Correct
D. 40 m/s
Solution: For the same direction the speeds subtract: v_AB = 25 − 15 = 10 m/s.
Q30 — Relative Motion in 2D · easy · numerical
Scooter A moves due north at 10 m/s and scooter B moves due east at 10 m/s. The magnitude of the velocity of A relative to B is:
A. 0 m/s
B. 10 m/s
C. 10√2 ≈ 14.1 m/s ✓ Correct
D. 20 m/s
Solution: v_AB = 10ĵ − 10î, so |v_AB| = √(10² + 10²) = 10√2 ≈ 14.1 m/s.