A rower on a river drops a bottle overboard and carries on upstream at a constant speed relative to the water for 20 minutes. He then turns and rows back downstream, catching the bottle at a point 2 km downstream of the drop. What is the speed of the current?

A rower on a river drops a bottle overboard and carries on upstream at a constant speed relative to the water for 20 minutes. He then turns and rows back downstream, catching the bottle at a point 2 km downstream of the drop. What is the speed of the current?

Approach: Switch to the frame moving with the water, where the bottle does not move at all and the rower's outward and return speeds are the same number.

3 km/h. In the frame of the water the bottle is stationary and the rower's relative motion is the same speed out and back, so the two legs take equal times and the return also lasts 20 minutes. That puts 40 minutes between the drop and the catch. During those 40 minutes the bottle drifts 2 km with the current, so the current is 2 km per 2/3 hour = 3 km/h. The rowing speed never enters the answer, which is why the problem looks underdetermined when set up in the ground frame. Doing it in the ground frame gives the same result once the rower's speed cancels from the two legs.

Follow-up: If the rower had turned after 20 minutes but rowed back at only half his earlier speed relative to the water, what would the current be?

Key concepts: frame of the water, relative motion, equal leg times.