Motion in a Plane – Relative Velocity in Two Dimensions: NEET MCQs with Answers

Explanation: The wind gives the rain a horizontal component of 12 m/s, on top of its vertical 35 m/s. The umbrella should be tilted at angle tan⁻¹(horizontal/vertical) = tan⁻¹(12/35) from the vertical, in the direction the rain (now slanted) is coming from.

Explanation: For the shortest possible path the boat must head partly upstream, so that the resultant velocity is directed straight across the river. The magnitude of this resultant is v_across = √(v_br² − v_r²) where v_br = 5 km/h is the boat's speed relative to the water and v_r the river speed. With width w = 1.0 km and time t = 15 min = 1/4 h: t = w/v_across ⇒ 1/4 = 1/√(25 − v_r²) √(25 − v_r²) = 4 25 − v_r² = 16 v_r² = 9 v_r = 3 km/h.

Explanation: To land exactly opposite the starting point, the resultant of the swimmer's velocity and the stream velocity must be perpendicular to the bank. This requires the component of the swimmer's velocity along the flow to cancel the stream speed u. The swimmer's velocity v makes 120° with the flow direction, so the angle it makes with the upstream direction is 180° − 120° = 60°, and its component opposing the flow is v cos 60° = v sin 30° Setting this equal to u: u = v sin 30° = 0.5 × (1/2) = 0.25 m/s. The remaining component, v sin 60° = 0.43 m/s, is what actually carries him across — which is why 0.43 appears as a distractor.

Explanation: Along the shortest path the boat travels straight across the river, so the upstream component of its velocity must cancel the current. The effective velocity carrying it across is therefore v_across = √(v_b² − u²) where v_b = 5 km/h is the boat's speed in still water and u the stream speed. The crossing takes t = 15 min = 1/4 h over a width d = 1 km, so v_across = d/t = 1/(1/4) = 4 km/h Hence √(25 − u²) = 4 25 − u² = 16 u² = 9 u = 3 km/h.

Explanation: Let C be the point where the boy from A catches the boy from B. In time t the boy from B has run a distance v₁t perpendicular to AB, while the boy from A has run a distance vt along the straight line AC. Since ABC is a right-angled triangle with the right angle at B: (vt)² = (v₁t)² + a² v²t² − v₁²t² = a² t²(v² − v₁²) = a² t = √[a²/(v² − v₁²)] = a/√(v² − v₁²). The expression makes physical sense only when v > v₁ — otherwise the boy at A can never catch up.

Explanation: Work in the frame of ship A. The velocity of B relative to A is v_BA = v_B − v_A = (10 km h⁻¹ North) − (10 km h⁻¹ West) = 10 km h⁻¹ North + 10 km h⁻¹ East So B approaches along the North-East direction, i.e. at 45°, with relative speed |v_BA| = √(10² + 10²) = 10√2 km h⁻¹ In this frame A is stationary and B travels along a straight line at 45° to the initial line joining them, which is 100 km long. The shortest distance is the perpendicular from A onto that line: PQ = 100 sin 45° = 100/√2 = 50√2 km The distance B covers along its relative path up to that point is also 100 cos 45° = 50√2 km, so t = 50√2/(10√2) = 5 h.

Explanation: The shortest path across a river is the straight line perpendicular to the banks. To achieve it, the swimmer must aim upstream so that the component of his own velocity along the flow exactly cancels the river current. If θ is the angle between his heading and the North direction, then v_SN sin θ = v_R sin θ = v_R/v_SN = 10/20 = 1/2 θ = 30° Since the river flows towards the East, he must angle himself towards the West to oppose it. Hence he should swim at 30° West of North. (Note this is different from the shortest-time crossing, for which he would simply head due North.)

Explanation: They collide only if the relative position and relative velocity are anti-parallel: (r₁−r₂)/|r₁−r₂| = (v₂−v₁)/|v₂−v₁|.

Explanation: Minimizing the separation d² = 200t² − 2000t + 10000 gives dd²/dt = 0 ⇒ t = 5 h.

Explanation: The boat's ground velocity is 1 m/s east plus 3 m/s north. The flag points along the wind velocity relative to the boat: (4 east) − (1 east + 3 north) = 3 m/s east and 3 m/s south. Equal east and south components mean the flag points south-east.

Explanation: Swimming straight across, the drift equals (river speed/swimmer speed) × width, so river speed/swimmer speed = 60/120 = 1/2. To land directly opposite, the upstream component of his velocity must cancel the current, so his heading makes angle φ with the perpendicular where sin φ = 1/2, φ = 30°. The angle with the bank is therefore 90° − 30° = 60°.

Explanation: The magnitude of the relative velocity is √(v₁² + v₂² − 2v₁v₂cos θ). It is largest when cos θ = −1, that is θ = π, giving 3 + 4 = 7 m/s.

Explanation: Heading straight across, the drift is (v_r/v_s) × width, so 60 = (v_r/v_s) × 100 and v_r/v_s = 0.6. To land directly opposite, the upstream component of his velocity must cancel the current: v_s sin θ = v_r, so sin θ = 0.6 and θ = 37° from the perpendicular.

Explanation: The velocity component across the river is 5 cos 37° = 4 m/s, so the crossing time is 120/4 = 30 s. Along the river, the swimmer moves 5 sin 37° = 3 m/s upstream relative to the water. The net velocity along the bank is 4 − 3 = 1 m/s downstream, so the drift is 1 × 30 = 30 m downstream.

Explanation: The crossing time depends only on the velocity component across the river, 5 × 0.8 = 4 m/s, so t = 120/4 = 30 s whatever the flow speed. The speed along the bank is now 6 + 5 × 0.6 = 9 m/s. The drift is therefore 9 × 30 = 270 m.