Q: The time to cross a river (by any heading) is decided by:

Only the component of the boat's velocity perpendicular to the banks. t = d/(v_br cosθ): only the across-river component matters; the current changes drift, never the crossing time.

Question 1

A boat crosses a river in the SHORTEST TIME when it is headed:

Accepted Answer

Straight across, perpendicular to the current. 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).

Question 2

The shortest time to cross a river of width d with boat speed v_br (relative to water) is:

Accepted Answer

t_{min} = \dfrac{d}{v_{br}}. 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.

Question 3

To cross the river along the SHORTEST PATH (reach the point directly opposite, zero drift), the boat must head:

Accepted Answer

Upstream at an angle to the perpendicular such that the current is cancelled. The upstream component v_br sinθ must cancel the current v_r: sinθ = v_r/v_br (θ measured from the straight-across direction).

Question 4

Crossing directly opposite (zero drift) is possible only if:

Accepted Answer

v_br > v_r. sinθ = v_r/v_br must be ≤ 1, so the boat speed must exceed the river speed. If v_r > v_br, some drift is unavoidable.

Question 5

For the shortest-path crossing, the resultant (ground) speed of the boat across the river is:

Accepted Answer

\sqrt{v_{br}^2 - v_r^2}. The upstream component cancels v_r; the across component left is √(v_br² − v_r²), so t = d/√(v_br² − v_r²).

Question 6

When a boat heads straight across a flowing river, its resultant velocity relative to the ground is:

Accepted Answer

\sqrt{v_{br}^2 + v_r^2}, directed at an angle downstream. The across and downstream components are perpendicular: v_ground = √(v_br² + v_r²), tilted downstream by tan⁻¹(v_r/v_br).

Question 7

The drift of a boat that heads straight across a river of width d is:

Accepted Answer

x = v_r \cdot \dfrac{d}{v_{br}}. Drift = (current speed) × (crossing time) = v_r · d/v_br. Speed-trick: drift = v_r t, with t from the perpendicular component only.

Question 8

A boat's speed downstream is ______ and upstream is ______ (v_b = boat speed in still water, v_r = current):

Accepted Answer

v_b + v_r ; v_b − v_r. Velocities add along the current (downstream) and subtract against it (upstream).

Question 9

If a swimmer's speed in still water is less than the river's speed, then to MINIMISE drift he should swim:

Accepted Answer

At an angle upstream such that cosθ = v_s/v_r (drift is minimised, not zero). Zero drift is impossible when v_s < v_r; the drift is minimised by heading upstream at the angle given by cosθ = v_s/v_r (θ from the upstream direction).

Question 10

The time to cross a river (by any heading) is decided by:

Accepted Answer

Only the component of the boat's velocity perpendicular to the banks. t = d/(v_br cosθ): only the across-river component matters; the current changes drift, never the crossing time.

Question 11

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:

Accepted Answer

20 s. t_min = d/v_br = 100/5 = 20 s.

Question 12

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:

Accepted Answer

90 m. t = 120/4 = 30 s; drift = v_r t = 3 × 30 = 90 m.

Question 13

A boat heads straight across a river at 4 m/s; the current is 3 m/s. The boat's resultant speed is:

Accepted Answer

5 m/s. v = √(4² + 3²) = √25 = 5 m/s (the classic 3-4-5 triangle).

Question 14

A swimmer can swim at 5 m/s in still water; the river flows at 3 m/s. To reach the point directly opposite, the swimmer's effective (ground) speed across is:

Accepted Answer

4 m/s. v_across = √(5² − 3²) = √16 = 4 m/s (3-4-5 again). Speed-trick: shortest path → subtract squares; shortest time → add squares.

Question 15

For the same swimmer (5 m/s, current 3 m/s) crossing a 200 m wide river along the shortest path, the time taken is:

Accepted Answer

50 s. t = d/√(v² − u²) = 200/4 = 50 s.

Question 16

A swimmer (speed 2 m/s) wants to cross directly opposite in a river flowing at 1 m/s. The angle (with the straight-across direction) at which he must head upstream is:

Accepted Answer

30°. sinθ = v_r/v_s = 1/2 ⇒ θ = 30° upstream of straight-across.

Question 17

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:

Accepted Answer

80 min. Downstream speed = 10 + 5 = 15 km/h; t = 20/15 h = 4/3 h = 80 min.

Question 18

The same boat (10 km/h still water, current 5 km/h) covers 20 km upstream in:

Accepted Answer

4 h. Upstream speed = 10 − 5 = 5 km/h; t = 20/5 = 4 h.

Question 19

A man swims at 4 km/h in still water across a river 1 km wide flowing at 3 km/h, heading straight across. His landing point is displaced downstream by:

Accepted Answer

0.75 km. t = 1/4 h; drift = 3 × 1/4 = 0.75 km.

Question 20

A boat crosses a 400 m river in minimum time 50 s and suffers a drift of 150 m. The speed of the current is:

Accepted Answer

3 m/s. v_br = 400/50 = 8 m/s (not needed); current = drift/time = 150/50 = 3 m/s.

River-Boat Problems — NEET Physics MCQs with Solutions

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