JEE Main 2026 — Motion In Two Dimensions Question with Solution

Width of the river $d=200$ m.
Speed of river $v_r=18$ km/h $=18\times \frac{5}{18}=5$ m/s.
Speed of boat in still water $v_b=36$ km/h $=36\times \frac{5}{18}=10$ m/s.
For minimum time to cross the river, the boat must head perpendicular to the river flow (at $90^{\circ }$ to the bank).
Time taken to cross once $t=\frac{d}{v_b}=\frac{200}{10}=20$ s.
For a round trip (bank to bank and back), the total minimum time is $T=2t=2\times 20=40$ s.
During the first crossing, the drift (displacement along the bank) is $x_1=v_r\times t=5\times 20=100$ 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 $x_2=v_r\times t=5\times 20=100$ m.
Total displacement along the river bank $=x_1+x_2=100+100=200$ m.
Thus, minimum time is $40$ s and displacement is $200$ m.