Learn › JEE Main · Physics › Motion in 2D
Motion in 2D — JEE Main Physics MCQs with Solutions
Free JEE Main 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.
▶ Practise Motion in 2D online (free)
Subtopics
Sample questions with solutions
Q1 — Basics of 2D · easy · numerical
The position of a particle is given by r = (8t î + 15t ĵ) m, where t is in seconds. What is the speed of the particle?
A. 23 m/s
B. 11.5 m/s
C. 7 m/s
D. 17 m/s ✓ Correct
Solution: v = dr/dt = (8 î + 15 ĵ) m/s, so speed = √(8² + 15²) = √289 = 17 m/s.
Q2 — Basics of 2D · easy · numerical
A particle moves so that its position is r = (7t î + 12t² ĵ) m. Find its speed at t = 1 s.
A. 19 m/s
B. 31 m/s
C. 25 m/s ✓ Correct
D. 13.9 m/s
Solution: v = (7 î + 24t ĵ); at t = 1 s, v = (7, 24), so speed = √(49 + 576) = 25 m/s.
Q3 — Basics of 2D · easy · numerical
The position of a particle is r = (4t³ î + 12t² ĵ) m. What is the magnitude of its acceleration at t = 1 s?
A. 48 m/s²
B. 12√2 ≈ 17.0 m/s²
C. 24√2 ≈ 33.9 m/s² ✓ Correct
D. 24 m/s²
Solution: a = d²r/dt² = (24t î + 24 ĵ); at t = 1 s, a = (24, 24), so |a| = 24√2 ≈ 33.9 m/s².
Q4 — Basics of 2D · easy · numerical
A car moving at 15 m/s due east is later found moving at 8 m/s due north, the change taking 2 s. Find the magnitude of its average acceleration.
A. 8.5 m/s² ✓ Correct
B. 11.5 m/s²
C. 3.5 m/s²
D. 17 m/s²
Solution: Δv = (0, 8) − (15, 0), so |Δv| = √(15² + 8²) = 17 m/s; average acceleration = 17/2 = 8.5 m/s².
Q5 — Basics of 2D · easy · numerical
A particle moves with x = 12t and y = 35t² (both in metres). Find the magnitude of its average velocity between t = 0 and t = 1 s.
A. 47 m/s
B. 23.5 m/s
C. 37 m/s ✓ Correct
D. 35 m/s
Solution: Displacement in 1 s is (12, 35) m, so |Δr| = √(144 + 1225) = 37 m and average velocity = 37/1 = 37 m/s.
Q6 — Basics of 2D · easy · numerical
A particle starts with velocity u = 20 î m/s and has constant acceleration a = 7 ĵ m/s². Find its speed at t = 3 s.
A. 29 m/s ✓ Correct
B. 21 m/s
C. 41 m/s
D. 27 m/s
Solution: v = u + at = (20, 0 + 7×3) = (20, 21) m/s, so speed = √(400 + 441) = √841 = 29 m/s.
Q7 — Projectile Motion · easy · numerical
A projectile is launched with velocity u = (20 î + 20 ĵ) m/s (g = 10 m/s²). Find its maximum height.
A. 40 m
B. 20 m ✓ Correct
C. 10 m
D. 80 m
Solution: H = u_y²/(2g) = 20²/(2×10) = 400/20 = 20 m.
Q8 — Projectile Motion · easy · numerical
A shell is fired at 20 m/s at 37° above the horizontal (sin 37° = 0.6, g = 10 m/s²). Find its time of flight.
A. 3.2 s
B. 4 s
C. 1.2 s
D. 2.4 s ✓ Correct
Solution: T = 2u sin θ/g = 2×20×0.6/10 = 2.4 s.
Q9 — Projectile Motion · easy · numerical
A ball is projected at 40 m/s at 45° to the horizontal (g = 10 m/s²). Find its horizontal range.
A. 160 m ✓ Correct
B. 320 m
C. 120 m
D. 80 m
Solution: R = u² sin 2θ/g = 1600 × sin 90°/10 = 1600/10 = 160 m.
Q10 — Projectile Motion · easy · numerical
A stone is thrown horizontally at 15 m/s from the top of a 20 m tower (g = 10 m/s²). How far from the base of the tower does it land?
A. 15 m
B. 30 m ✓ Correct
C. 40 m
D. 60 m
Solution: Fall time t = √(2h/g) = √(2×20/10) = 2 s, so horizontal distance = 15×2 = 30 m.
Q11 — Projectile Motion · easy · numerical
A ball is thrown horizontally at 8 m/s from a tall building (g = 10 m/s²). Find its speed 1.5 s after it is thrown.
A. 23 m/s
B. 15 m/s
C. 12.5 m/s
D. 17 m/s ✓ Correct
Solution: After 1.5 s, vₓ = 8 m/s and v_y = gt = 10×1.5 = 15 m/s, so speed = √(64 + 225) = 17 m/s.
Q12 — Projectile Motion · easy · numerical
A projectile is fired at 60 m/s at 30° above the horizontal (g = 10 m/s²). Find its maximum height.
A. 45 m ✓ Correct
B. 30 m
C. 90 m
D. 180 m
Solution: u_y = 60 sin 30° = 30 m/s, so H = u_y²/(2g) = 900/20 = 45 m.
Q13 — Relative Motion in 2D · easy · numerical
Car A travels due east at 21 m/s while car B travels due north at 20 m/s. Find the magnitude of the velocity of A relative to B.
A. 20.5 m/s
B. 41 m/s
C. 29 m/s ✓ Correct
D. 1 m/s
Solution: v_AB = v_A − v_B = (21, 0) − (0, 20) = (21, −20) m/s, so |v_AB| = √(441 + 400) = 29 m/s.
Q14 — Relative Motion in 2D · easy · numerical
To a cyclist riding due east at 12 m/s, a bird appears to fly due north at 35 m/s. Find the true speed of the bird.
A. 47 m/s
B. 37 m/s ✓ Correct
C. 35 m/s
D. 23 m/s
Solution: v_bird = v_cyclist + v_bird/cyclist = (12, 0) + (0, 35) = (12, 35) m/s, so true speed = √(144 + 1225) = 37 m/s.
Q15 — Relative Motion in 2D · easy · numerical
Two particles leave a common point, each moving at 8 m/s, along straight lines 60° apart. Find the magnitude of the velocity of one relative to the other.
A. 8 m/s ✓ Correct
B. 4 m/s
C. 8√3 ≈ 13.9 m/s
D. 16 m/s
Solution: |v_rel| = 2v sin(θ/2) = 2×8×sin 30° = 8 m/s.
Q16 — Relative Motion in 2D · easy · numerical
Two trains leave a junction at the same moment, each at 7 m/s, along two perpendicular straight tracks. How far apart are they after 10 s?
A. 140 m
B. 49 m
C. 70 m
D. 70√2 ≈ 99 m ✓ Correct
Solution: After 10 s each has gone 70 m along perpendicular directions, so separation = √(70² + 70²) = 70√2 ≈ 99 m.
Q17 — Relative Motion in 2D · easy · numerical
Particle A has velocity (8 î + 4 ĵ) m/s and particle B has velocity (−7 î + 4 ĵ) m/s. Find the speed of A relative to B.
A. 1 m/s
B. 17 m/s
C. 8 m/s
D. 15 m/s ✓ Correct
Solution: v_AB = (8 − (−7), 4 − 4) = (15, 0) m/s, so relative speed = 15 m/s.
Q18 — Relative Motion in 2D · easy · numerical
One drone flies toward the northeast at 10√2 m/s while another flies due north at 10 m/s. Find the speed of the first drone relative to the second.
A. 20 m/s
B. 10√2 ≈ 14.1 m/s
C. 10 m/s ✓ Correct
D. 5 m/s
Solution: The first drone's velocity is (10, 10) m/s and the second's is (0, 10) m/s, so v_rel = (10, 0) and the relative speed is 10 m/s.
Q19 — Rain Man Problem · easy · numerical
Rain is falling vertically with a speed of 15 m/s. A man runs horizontally at 8 m/s on a straight road. What is the speed of the rain relative to the man?
A. 12 m/s
B. 15 m/s
C. 23 m/s
D. 17 m/s ✓ Correct
Solution: Relative to the man the rain has a vertical component 15 m/s and a horizontal component 8 m/s, so v = √(15² + 8²) = √289 = 17 m/s.
Q20 — Rain Man Problem · easy · numerical
Rain falls vertically at 24 m/s. To a cyclist riding at 7 m/s on a level road, at what angle with the vertical does the rain appear to fall?
A. tan⁻¹(24/25)
B. tan⁻¹(7/25)
C. tan⁻¹(7/24) ✓ Correct
D. tan⁻¹(24/7)
Solution: In the cyclist frame the rain gains a horizontal component 7 m/s while keeping its vertical component 24 m/s, so tan θ = 7/24 ⇒ θ = tan⁻¹(7/24) from the vertical.
Q21 — Rain Man Problem · easy · numerical
A car is moving at 24 m/s on a straight road while rain falls vertically at 7 m/s. With what speed do the raindrops strike the windscreen (speed of rain relative to the car)?
A. 17 m/s
B. 25 m/s ✓ Correct
C. 24 m/s
D. 31 m/s
Solution: Speed of rain relative to the car = √(24² + 7²) = √(576 + 49) = √625 = 25 m/s.
Q22 — Rain Man Problem · easy · numerical
To a driver moving at 15 m/s, vertically falling rain appears to have a speed of 17 m/s. What is the actual speed of the rain?
A. 32 m/s
B. 2 m/s
C. 8 m/s ✓ Correct
D. 11 m/s
Solution: Apparent speed² = actual speed² + car speed², so actual speed = √(17² − 15²) = √(289 − 225) = √64 = 8 m/s.
Q23 — Rain Man Problem · easy · numerical
Rain falls vertically at 15 km/h. To a person moving on a level road the rain appears to make an angle tan⁻¹(8/15) with the vertical. Find the speed of the person.
A. 15 km/h
B. 4 km/h
C. 17 km/h
D. 8 km/h ✓ Correct
Solution: tan θ = v_person/v_rain, so v_person = 15 × (8/15) = 8 km/h.
Q24 — Rain Man Problem · easy · numerical
Rain is falling vertically. To a motorist driving at 20 m/s the drops appear to come at 45° with the vertical. The actual speed of the rain is:
A. 10 m/s
B. 20 m/s ✓ Correct
C. 20√2 ≈ 28 m/s
D. 14 m/s
Solution: tan 45° = v_car/v_rain = 1, so v_rain = 20 m/s.
Q25 — River Boat Problem · easy · numerical
A river is 240 m wide. A boat whose speed in still water is 8 m/s heads perpendicular to the current, which flows at 5 m/s. How long does the boat take to reach the opposite bank?
A. 48 s
B. 40 s
C. 25 s
D. 30 s ✓ Correct
Solution: The current does not affect the motion across the river: t = width/boat speed = 240/8 = 30 s.
Q26 — River Boat Problem · easy · numerical
A boat with still-water speed 8 m/s heads perpendicular to the banks of a 200 m wide river flowing at 15 m/s. How far downstream from the point directly opposite does it land?
A. 300 m
B. 375 m ✓ Correct
C. 160 m
D. 200 m
Solution: Crossing time t = 200/8 = 25 s; drift = river speed × time = 15 × 25 = 375 m.
Q27 — River Boat Problem · easy · numerical
A boat can move at 20 km/h in still water. It heads straight across a river flowing at 21 km/h. The resultant speed of the boat is:
A. 41 km/h
B. 21 km/h
C. 29 km/h ✓ Correct
D. 24.5 km/h
Solution: The two velocities are perpendicular: v = √(20² + 21²) = √(400 + 441) = √841 = 29 km/h.
Q28 — River Boat Problem · easy · numerical
A motorboat can travel at 24 km/h in still water. The river flows at 7 km/h. How long does the boat take to go 62 km downstream?
A. 2 h ✓ Correct
B. 1.5 h
C. 3.6 h
D. 2.6 h
Solution: Downstream speed = 24 + 7 = 31 km/h; t = 62/31 = 2 h.
Q29 — River Boat Problem · easy · numerical
A boat whose speed in still water is 20 km/h has to travel 36 km upstream in a river flowing at 8 km/h. The time taken is:
A. 2.5 h
B. 1.8 h
C. 1.3 h
D. 3 h ✓ Correct
Solution: Upstream speed = 20 − 8 = 12 km/h; t = 36/12 = 3 h.
Q30 — River Boat Problem · easy · numerical
A boat can move at 17 km/h in still water in a river flowing at 8 km/h. If it crosses along the shortest path (straight across), its resultant speed is:
A. 9 km/h
B. 17 km/h
C. 15 km/h ✓ Correct
D. 25 km/h
Solution: For the shortest path the upstream heading component cancels the current, so v = √(17² − 8²) = √225 = 15 km/h.