Two Swimmers

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Problem:

Two swimmers leave point $A$ on one bank of the river to reach point $B$ lying right across on the other bank. One of them crosses the river along the straight line $AB$ while the other swims at right angles to the stream and then walks the distance that he has been carried away by the stream to get to point $B$. What was the velocity $u$ of his walking if both swimmers reached the destination simultaneously? The stream velocity $v_0 = 2.0 \text{ km/hour}$ and the velocity if of each swimmer with respect to water equals $2.5 \text{ km per hour}$. 


Solution:

WLOG assume that the flow of the river (aka the stream) is towards the right. Let $v$ be the final vector from $A$ to $B$. 

First consider the first person. To swim directly from $A$ to $B$, the person must work diagonally for vector $v_p$. Simultaneously, the stream will drag the person along the vector $v_s$. These two result in the final vector $v_p+v_s=v$. So, he takes $$t_1=\frac{AB}{\sqrt{v_p^2-v_s^2}}$$ hours to get to point $B$. 


For the second swimmer, note that he takes $t_2=\frac{AB}{v}$ hours to get to point $C$, which is the point that he ends up after swimming. The distance he has drifted by the stream, namely $BC$, is $v_st_2=v_s(\frac{AB}{v})$. Then, the time this swimmer takes to walk from $C$ to $B$ is $\frac{x}{u}=\frac{v_0(AB)}{vu}$. Hence the total time this swimmer takes, swimming and walking, is $$\frac{AB}{v}+\frac{v_0(AB)}{vu}.$$


Now the problem tells us that $t_1=t_2+t_3$, so $$\frac{AB}{\sqrt{v_p^2-v_s^2}}=\frac{AB}{v}+\frac{v_0(AB)}{vu}$$ $$\implies u=\frac{v_s}{(\frac{1-v_s^2}{v^2})^\frac{-1}{2}}-1$$ $$=\boxed{3}.$$

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