Physics · Motion In A Plane · NEET
No, and this is the most tested trap. Shortest TIME means you cross in the least seconds. You get this by swimming straight across (perpendicular to the bank), because then your entire speed v_swim works to close the width d. The river still carries you downstream, so your actual path is slanted and longer, but the time is minimum: t = d / v_swim. Shortest PATH means the straight line across, with zero downstream drift. To do that you must aim upstream so the current is cancelled, but then only part of your speed crosses the river, so it takes MORE time. You cannot have both at once, unless the river is still.
The river current pushes you downstream. To land exactly opposite your start point, you must create an equal and opposite upstream velocity component. If you aim upstream at angle theta from the perpendicular, your upstream component is v_swim sin(theta). Setting this equal to the current v_river gives v_swim sin(theta) = v_river, so sin(theta) = v_river / v_swim. Only the remaining component v_swim cos(theta) actually carries you across the width.
Aim straight across, perpendicular to the bank. Time to cross depends only on the velocity component perpendicular to the bank. That component is largest (equal to full v_swim) when you point straight across. Any upstream or downstream tilt reduces the perpendicular component and increases the time. So shortest time = swim perpendicular, and t_min = d / v_swim, no matter what the current is.
For the straight-across path you only use the perpendicular component v_swim cos(theta). So the time is t = d / (v_swim cos(theta)), which is larger than d / v_swim because cos(theta) is less than 1. Example: if v_swim = 20 and v_river = 10, then sin(theta) = 0.5, so theta = 30 degrees, cos(30) = 0.866, and the shortest-path time is about 15 percent longer than the shortest-time crossing.
If v_river is greater than v_swim, then v_river / v_swim is greater than 1, and sin(theta) cannot exceed 1. This means it is impossible to cross by the shortest (straight-across) path; the swimmer will always drift downstream and land at some angle. Minimum drift is then achieved at a specific aiming angle, but reaching the exact opposite point is not possible.
The speed of a swimmer in still water is 20 m/s. The speed of river water is 10 m/s flowing due east. If he is standing on the south bank and wishes to cross the river along the shortest path, the angle at which he should make his strokes with respect to north is:
Try the real previous-year questions from this chapter — each with the answer and a full solution.
sin(theta) = v_river / v_swim, where theta is measured from the perpendicular (the straight-across direction) toward upstream. The swimmer aims upstream by this angle so the current is exactly cancelled and the net path is straight across.
t_min = d / v_swim. You get this by swimming perpendicular to the bank so your full speed carries you across. The current does not change this time because it only moves you along the bank, not across it.
No. The crossing time depends only on the velocity component perpendicular to the bank. The current is parallel to the bank, so it changes where you land (the drift), but not how long the crossing takes when you aim straight across.
Time to cross is t = d / v_swim. During that time the current carries you a distance x = v_river * t = v_river * d / v_swim along the bank. That sideways distance x is the drift.
Yes. River-boat and swimmer problems are a direct application of relative velocity in two dimensions from Motion in a Plane, and NEET has asked it (for example NEET 2019). The shortest-path vs shortest-time distinction is a favourite trap, so it is high-value to master.
