Question

boat #3 travels directly across the river but the river current has increased to 10 m/s because of increased rain and the boat moves at its top speed, 4 m/s.
how will the distance the boat #3 moves downstream during the trip compare to boat #1?
a 1/4 the distance
b 1/2 the distance
c the same distance

Explanation:

Step1: Calculate the time taken to cross the river

The time \(t\) taken by a boat to cross a river of width \(d\) with a velocity \(v_y\) (perpendicular to the flow) is given by \(t=\frac{d}{v_y}\). Since the boat's velocity perpendicular to the river (assuming it's moving straight across as in typical river - crossing problems, and here the boat's speed across the river is its own speed, say \(v = 4m/s\)) is the same for both cases (the river width \(d\) is constant and the boat's speed across the river \(v\) is constant). Let the width of the river be \(d\). For Boat #1 (original river speed, say \(v_{r1}\)) and Boat #3 (new river speed \(v_{r2}=10m/s\)), the time to cross the river \(t=\frac{d}{4}\) (because the motion across the river is independent of the river's current. The river's current affects the downstream motion, not the cross - river motion).

Step2: Calculate the downstream distance

The downstream distance \(x\) is given by \(x = v_{r}\times t\). Since \(t=\frac{d}{4}\) (constant) and \(v_{r}\) is the river's speed. For Boat #1, let the river speed be \(v_{r1}\), and for Boat #3, \(v_{r2}\). But wait, no, actually, if we assume that in the first case (Boat #1), maybe there was a miscalculation. Wait, no, the key is that the time to cross the river \(t\) (which determines the downstream drift) is set by the boat's speed across the river. The formula for the downstream distance \(x\) (due to the river current) is \(x=v_{r}\times t\), where \(t = \frac{d}{v_{boat - across}}\) (\(v_{boat - across}=4m/s\) is constant). The river current \(v_{r}\) doesn't affect the time to cross the river. So the time \(t\) to cross the river is the same for both boats (because \(t=\frac{d}{v_{boat - across}}\), \(d\) (width of river) and \(v_{boat - across}=4m/s\) are constant).

Answer:

C. The same distance