Question

2. Alice and Bob are swimming in a river. They start together at the same point on the bank of a wide stream that flows with a speed v. Both move at the same speed c where c> v) relative to the water. Alice swims downstream a distance L and then upstream the same distance. Bob swims so that his motion relative to the Earth is perpendicular to the banks of the stream. He swims the distance L and then back the same distance, with both swimmers returning to the starting point. In terms of L,c, and v, find the time intervals required (a) for Alice's round trip. (b) for Bob's round trip (c) Explain which swimmer returns first. (Refer back to this calculation in physics 4C and 4D when you consider the Michelson

Answer 1

Alice

Speed w.r.t. ground frame for the first part of the trip

$c+v$

The corresponding time taken is

$t_1=\frac{L}{c+v}$

similarly, for the second part of the trip

$t_2=\frac{L}{c-v}$

So the total time is

$t_A=\frac{2cL}{c^2-v^2}$

(b)Bob

If the swimmer is to flow perpendicular to the ground frame, the $\theta$ must be such

$c\sin\theta=v\Rightarrow\sin\theta=\frac{v}{c}$

So that the speed of swimmer in the ground frame is

$c\cos\theta=\sqrt{c^2-v^2}$

So the time of travel is

$t_1=\frac{L}{\sqrt{c^2-v^2}}$

For the second part of trip, the situation is very much the same.

$t_2=t_1=\frac{L}{\sqrt{c^2-v^2}}$

So that the total time is

$t_B=\frac{2L}{\sqrt{c^2-v^2}}$

(c) Compare

$t_A \textup{ and }t_B$

$\frac{c}{c^2-v^2}\textup{ and }\frac{1}{\sqrt{c^2-v^2}}$

$c\textup{ and }\sqrt{c^2-v^2}$

$c^2\textup{ and }c^2-v^2$

Thus, A's time is greater so B reaches first.