Kinematics

JEE Main level, single correct, medium

182 of 308

Question 182 of 308

A river of width 200 m is flowing from west to east with a speed of 18 km/h18 \mathrm{~km} / \mathrm{h}. A boat, moving with speed of 36 km/h36 \mathrm{~km} / \mathrm{h} in still water, is made to travel one-round trip (bank to bank of the river). Minimum time taken by the boat for this journey and also the displacement along the river bank are ____\_\_\_\_ and ____\_\_\_\_ respectively.

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Answer: (C) 40 s and 200 m

Width of the river d=200d = 200 m.
Speed of river vr=18v_r = 18 km/h =18×518=5= 18 \times \frac{5}{18} = 5 m/s.
Speed of boat in still water vb=36v_b = 36 km/h =36×518=10= 36 \times \frac{5}{18} = 10 m/s.
For minimum time to cross the river, the boat must head perpendicular to the river flow (at 90∘90^\circ to the bank).
Time taken to cross once t=dvb=20010=20t = \frac{d}{v_b} = \frac{200}{10} = 20 s.
For a round trip (bank to bank and back), the total minimum time is T=2t=2×20=40T = 2t = 2 \times 20 = 40 s.
During the first crossing, the drift (displacement along the bank) is x1=vr×t=5×20=100x_1 = v_r \times t = 5 \times 20 = 100 m.
During the return trip, to minimize time, the boat again heads perpendicular to the flow relative to the water. The river continues to flow in the same direction, so the drift is again in the same direction as the first trip.
Return drift x2=vr×t=5×20=100x_2 = v_r \times t = 5 \times 20 = 100 m.
Total displacement along the river bank =x1+x2=100+100=200= x_1 + x_2 = 100 + 100 = 200 m.
Thus, minimum time is 4040 s and displacement is 200200 m.

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