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 and B is 1200 m. The velocity of the river current `v= 1.9 m//s` is constant over the entire width of the river. The line AB makes an angle `alpha= 60^@` 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

Correct Answer - A::B
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.

`u sin beta = v sin alpha` ...(i)
`S = AB = (u cos beta + v cos alpha) t_1` ...(ii)
Further `BA= (u cos beta - v cos alpha) t_2` ...(iii)
`t_1 + t_2 = t` ...(iv)
Solving these equations after proper substitution,
we get
`u= 8 m//s and beta=12^@`