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.

Options

A. The velocity of the boat is `8m//s`
B. The velocity of the boat is `6m//s`
C. The angle made by `u` with the line `AB` is `12^(@)`
D. The angle made by `u` with the line `AB` is `10^(@)`

Answer

A, C

Explanation

In the 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`
Solving these equations after proper substitution,
we get `u=8m//s and beta=12^(0)`.