Question

A man wants to swim in a river from $A$ to $B$ and back from $B$ to $A$ always following line $AB$. The distance between points $A$ and $B$ is S. The velocity of the river current v is constant over the entire width of the river. $A$ The line $AB$ makes an angle $\alpha$ with the direction of current. With what velocity u and at what angle $\beta$ to the line $AB$ should the man swim to cover distance $AB$ and back in time $t$?

Answer 1

In this problem, we choose axis along $AB$ and normal to it.
Since the man moves along $AB$, the velocity components of current and man must cancel out, i.e.,
$v\sin \beta =v\sin \alpha$
When the man moves from A to B, his resultant velocity along
$AB=(u\cos \beta +v\cos \alpha )$
Hence, $S=(u\cos \beta +v\cos \alpha )t_1$
while for motion from B to $AS=(u\cos \beta -v\cos \alpha )t_2$
From the condition of the problem, $t_1+t_2=t$
$\frac{S}{u\cos \beta +v\cos \alpha }+\frac{S}{u\cos \beta -v\cos \alpha }=t$
$S\left[\frac{u\cos \beta -v\cos \alpha +u\cos \beta +v\cos \alpha }{u^2{\cos }^2\beta +v^2{\cos }^2\alpha }\right]=t$
$\frac{S\left(2u\cos \beta \right)}{u^2{\cos }^2\beta +v^2{\cos }^2\alpha }=t$