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=19 m/s is constant over the entire width of the river. The line AB makes an angle α=60o with the direction of the current. With what velocity u and at what angle β to the line AB should the launch move to cover the distance AB and back in a time t=5 min? The angle β remains the same during the passage from A to B and from B to A.

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Solution

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.,

usinβ=vsinα ....................................................(i)

S=AB=(ucosβ−vcosα)t−1 ....................................(ii)

Further BA=(ucosβ−vcosα)t2 ......................................(iii)

t1+t2=t .........................................................................(iv)

Solving these equations after proper substitution, we get

u=8m/s and β=12o

usinβ=vsinα ....................................................(i)

S=AB=(ucosβ−vcosα)t−1 ....................................(ii)

Further BA=(ucosβ−vcosα)t2 ......................................(iii)

t1+t2=t .........................................................................(iv)

Solving these equations after proper substitution, we get

u=8m/s and β=12o

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