Question

A launch plies between two points $A$ and $B$ on the opposite banks of a river always following the line $AB$. The distance $S$ between points $A$ and $B$ is $1200$ m. the velocity of the river current $v=\frac{1}{9}$ m/s is constant over the entire width of the river. The line $AB$ makes an angle $\alpha ={60}^o$ with the direction of the current. With what velocity $u$ and at what angle $\beta$ to the line $AB$ should the launch move to cover the distance $AB$ and back in a time $t=5$ min? The angle $\beta$ remains the same during the passage from $A$ to $B$ and from $B$ to $A$.

Answer 1

In order that the moving launch is always on the straight line AB, the components of velocity of the current and of the launch in the direction perpendicular to AB should be equal, i.e.,
$usin\beta =vsin\alpha$ ....................................................(i)
$S=AB=(ucos\beta -vcos\alpha )t-1$ ....................................(ii)
Further $BA=(ucos\beta -vcos\alpha )t_2$ ......................................(iii)
$t_1+t_2=t$ .........................................................................(iv)
Solving these equations after proper substitution, we get
$u=8m/s$ and $\beta ={12}^o$