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Relative Motion (1D and 2D) — IISER Physics MCQs with Solutions
Free IISER Physics Relative Motion (1D and 2D) MCQs with step-by-step solutions covering Relative Motion in 1D, River-Boat Problems, Rain-Man Problems, Closest Approach & Wind-Airplane Problems. Practise online on Prepizo — no login needed.
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Sample questions with solutions
Q1 — Relative Motion in 1D · easy · theory
The velocity of a body A relative to a body B is defined as:
A. $\vec{v}_{AB} = \vec{v}_A - \vec{v}_B$ ✓ Correct
B. $\vec{v}_{AB} = \vec{v}_A \times \vec{v}_B$
C. $\vec{v}_{AB} = \vec{v}_A + \vec{v}_B$
D. $\vec{v}_{AB} = \vec{v}_B - \vec{v}_A$
Solution: Core equation: v⃗_AB = v⃗_A − v⃗_B — the velocity of A as measured in the frame of B.
Q2 — Relative Motion in 1D · easy · theory
Two cars move in the SAME direction with speeds v_A and v_B (v_A > v_B). The magnitude of the relative velocity of A with respect to B is:
A. v_A
B. zero
C. v_A + v_B
D. v_A − v_B ✓ Correct
Solution: Same direction ⇒ magnitudes subtract: |v_AB| = v_A − v_B. Speed-trick: same direction → subtract; opposite → add.
Q3 — Relative Motion in 1D · easy · theory
Two trains approach each other with speeds v₁ and v₂. The relative speed of one with respect to the other is:
A. v₁ + v₂ ✓ Correct
B. √(v₁² + v₂²)
C. zero
D. v₁ − v₂
Solution: Opposite directions ⇒ relative speed = v₁ + v₂ (they close the gap at the sum of speeds).
Q4 — Relative Motion in 1D · easy · theory
The relative velocity of A with respect to B and the relative velocity of B with respect to A are:
A. Unrelated to each other
B. Equal in magnitude but opposite in direction ✓ Correct
C. Equal in magnitude and direction
D. Always zero
Solution: v⃗_AB = −v⃗_BA. They have the same magnitude and opposite directions.
Q5 — Relative Motion in 1D · easy · theory
Two bodies moving with equal velocities (same speed and direction) have a relative velocity of:
A. Infinite
B. Equal to their common speed
C. Twice the common speed
D. Zero — each appears at rest to the other ✓ Correct
Solution: v⃗_AB = v⃗_A − v⃗_B = 0; each body appears stationary in the other's frame (like two parallel trains at equal speed).
Q6 — Relative Motion in 1D · easy · theory
When a train passes a pole, the distance it covers (relative to the pole) equals:
A. Its own length ✓ Correct
B. The length of the platform
C. Zero
D. Twice its length
Solution: To pass a point object the train must move forward by its own length L, so t = L/v.
Q7 — Relative Motion in 1D · easy · numerical
Car A moves at 60 km/h and car B at 40 km/h in the same direction. The velocity of A relative to B is:
A. 20 km/h ✓ Correct
B. 100 km/h
C. 40 km/h
D. 60 km/h
Solution: v_AB = v_A − v_B = 60 − 40 = 20 km/h (in the direction of motion).
Q8 — Relative Motion in 1D · easy · numerical
Two trains approach each other at 40 km/h and 50 km/h. Their relative speed is:
A. 45 km/h
B. 90 km/h ✓ Correct
C. 2000 km/h
D. 10 km/h
Solution: Opposite directions ⇒ v_rel = 40 + 50 = 90 km/h.
Q9 — Relative Motion in 1D · easy · numerical
A 100 m long train moving at 20 m/s crosses a pole in:
A. 10 s
B. 2 s
C. 5 s ✓ Correct
D. 20 s
Solution: t = L/v = 100/20 = 5 s.
Q10 — Relative Motion in 1D · easy · numerical
Two cars, 150 m apart, drive towards each other at 10 m/s and 5 m/s. They meet after:
A. 10 s ✓ Correct
B. 30 s
C. 15 s
D. 7.5 s
Solution: t = separation/relative speed = 150/(10 + 5) = 10 s.
Q11 — River-Boat Problems · easy · theory
A boat crosses a river in the SHORTEST TIME when it is headed:
A. Straight across, perpendicular to the current ✓ Correct
B. Downstream at an angle
C. Directly against the current
D. Upstream at an angle
Solution: Minimum time needs the full boat speed across the river: head perpendicular to the bank. t_min = d/v_br (the drift is then unavoidable).
Q12 — River-Boat Problems · easy · theory
The shortest time to cross a river of width d with boat speed v_br (relative to water) is:
A. $t_{min} = \dfrac{d}{v_{br} + v_r}$
B. $t_{min} = \dfrac{d}{\sqrt{v_{br}^2 - v_r^2}}$
C. $t_{min} = \dfrac{d}{v_{br} - v_r}$
D. $t_{min} = \dfrac{d}{v_{br}}$ ✓ Correct
Solution: Crossing time depends only on the perpendicular component; at full v_br across, t_min = d/v_br. The current does not affect crossing time.
Q13 — River-Boat Problems · easy · theory
To cross the river along the SHORTEST PATH (reach the point directly opposite, zero drift), the boat must head:
A. Along the current
B. Downstream at an angle
C. Straight across
D. Upstream at an angle to the perpendicular such that the current is cancelled ✓ Correct
Solution: The upstream component v_br sinθ must cancel the current v_r: sinθ = v_r/v_br (θ measured from the straight-across direction).
Q14 — River-Boat Problems · easy · theory
When a boat heads straight across a flowing river, its resultant velocity relative to the ground is:
A. v_br − v_r, straight across
B. v_r, along the current
C. v_br, straight across
D. $\sqrt{v_{br}^2 + v_r^2}$, directed at an angle downstream ✓ Correct
Solution: The across and downstream components are perpendicular: v_ground = √(v_br² + v_r²), tilted downstream by tan⁻¹(v_r/v_br).
Q15 — River-Boat Problems · easy · theory
A boat's speed downstream is ______ and upstream is ______ (v_b = boat speed in still water, v_r = current):
A. v_b − v_r ; v_b + v_r
B. v_b ; v_b
C. v_r ; v_b
D. v_b + v_r ; v_b − v_r ✓ Correct
Solution: Velocities add along the current (downstream) and subtract against it (upstream).
Q16 — River-Boat Problems · easy · numerical
A river is 100 m wide. A boat with speed 5 m/s (relative to water) heads straight across. The minimum time to cross is:
A. 50 s
B. 25 s
C. 10 s
D. 20 s ✓ Correct
Solution: t_min = d/v_br = 100/5 = 20 s.
Q17 — River-Boat Problems · easy · numerical
A boat heads straight across a 120 m wide river at 4 m/s while the current flows at 3 m/s. The drift when it reaches the other side is:
A. 90 m ✓ Correct
B. 40 m
C. 120 m
D. 160 m
Solution: t = 120/4 = 30 s; drift = v_r t = 3 × 30 = 90 m.
Q18 — River-Boat Problems · easy · numerical
A boat heads straight across a river at 4 m/s; the current is 3 m/s. The boat's resultant speed is:
A. 3.5 m/s
B. 5 m/s ✓ Correct
C. 7 m/s
D. 1 m/s
Solution: v = √(4² + 3²) = √25 = 5 m/s (the classic 3-4-5 triangle).
Q19 — River-Boat Problems · easy · numerical
A boat's speed in still water is 10 km/h. It travels 20 km downstream in a river flowing at 5 km/h in:
A. 2 h
B. 4 h
C. 80 min ✓ Correct
D. 60 min
Solution: Downstream speed = 10 + 5 = 15 km/h; t = 20/15 h = 4/3 h = 80 min.
Q20 — River-Boat Problems · easy · numerical
The same boat (10 km/h still water, current 5 km/h) covers 20 km upstream in:
A. 1.5 h
B. 2 h
C. 80 min
D. 4 h ✓ Correct
Solution: Upstream speed = 10 − 5 = 5 km/h; t = 20/5 = 4 h.
Q21 — Rain-Man Problems · easy · theory
The velocity of rain relative to a moving man is given by:
A. $\vec{v}_{rm} = \vec{v}_r \times \vec{v}_m$
B. $\vec{v}_{rm} = \vec{v}_r + \vec{v}_m$
C. $\vec{v}_{rm} = \vec{v}_r - \vec{v}_m$ ✓ Correct
D. $\vec{v}_{rm} = \vec{v}_m - \vec{v}_r$
Solution: Core equation: v⃗_rm = v⃗_rain − v⃗_man. The umbrella must be held along v⃗_rm.
Q22 — Rain-Man Problems · easy · theory
Rain falls vertically. A man walking forward should hold his umbrella:
A. Tilted forward (towards the direction of his motion) ✓ Correct
B. Vertically
C. Horizontally
D. Tilted backward
Solution: Relative to the man, the rain acquires a backward horizontal component (−v⃗_m), so it appears to come from the front — tilt the umbrella forward.
Q23 — Rain-Man Problems · easy · theory
Rain falls vertically at speed v_r while a man walks at v_m. The angle of the umbrella from the vertical should satisfy:
A. $\sin\theta = \dfrac{v_r}{v_m}$
B. $\cos\theta = \dfrac{v_m}{v_r}$
C. $\tan\theta = \dfrac{v_m}{v_r}$ ✓ Correct
D. $\tan\theta = \dfrac{v_r}{v_m}$
Solution: The apparent rain has horizontal component v_m and vertical component v_r: tanθ = v_m/v_r (θ from the vertical, tilted towards motion). Speed-trick: faster walking → bigger tilt.
Q24 — Rain-Man Problems · easy · theory
If the man starts running FASTER (rain still vertical), the umbrella angle from the vertical:
A. Decreases
B. Stays the same
C. Becomes zero
D. Increases ✓ Correct
Solution: tanθ = v_m/v_r grows with v_m: the faster he runs, the more he must tilt the umbrella forward.
Q25 — Rain-Man Problems · easy · theory
To a man moving in a car, vertically falling rain strikes the windscreen obliquely. The faster the car moves, the rain appears to come:
A. From directly above always
B. From behind
C. More horizontally (from the front) ✓ Correct
D. More vertically
Solution: The backward relative component grows with car speed, so the apparent rain direction tilts towards the horizontal front.
Q26 — Rain-Man Problems · easy · numerical
Rain falls vertically at 10 m/s while a man walks at 10 m/s. He must hold his umbrella at an angle from the vertical of:
A. 30°
B. 45° ✓ Correct
C. 90°
D. 60°
Solution: tanθ = v_m/v_r = 10/10 = 1 ⇒ θ = 45° (tilted forward).
Q27 — Rain-Man Problems · easy · numerical
Rain falls vertically at 30 m/s; a car moves at 10 m/s. The rain strikes the windscreen at an angle from the vertical of:
A. tan⁻¹(3) ≈ 71.6°
B. 45°
C. tan⁻¹(1/3) ≈ 18.4° ✓ Correct
D. 30°
Solution: tanθ = v_car/v_rain = 10/30 = 1/3 ⇒ θ = tan⁻¹(1/3) ≈ 18.4°.
Q28 — Rain-Man Problems · easy · numerical
Rain falls vertically at 4 m/s; a man walks at 3 m/s. The speed of the rain relative to the man is:
A. 5 m/s ✓ Correct
B. 7 m/s
C. 3.5 m/s
D. 1 m/s
Solution: |v_rm| = √(4² + 3²) = 5 m/s (3-4-5 triangle).
Q29 — Rain-Man Problems · easy · numerical
Vertically falling rain has speed 12 m/s; a man runs at 5 m/s. The relative speed of the rain with respect to him is:
A. 7 m/s
B. 12 m/s
C. 17 m/s
D. 13 m/s ✓ Correct
Solution: |v_rm| = √(12² + 5²) = √169 = 13 m/s (5-12-13 triple).
Q30 — Closest Approach & Wind-Airplane Problems · easy · theory
In the reference frame of particle B, particle A (both moving uniformly) appears to move:
A. In a straight line with constant velocity v⃗_A − v⃗_B ✓ Correct
B. At rest always
C. In a parabola
D. In a circle
Solution: With both velocities constant, the relative velocity is constant, so the relative trajectory is a straight line.