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Question

A man wants to swim across a river from A to B and back from B to A always following line AB. The distance S between points A and B is S. The velocity of the river current v is constant over the entire width of the river. The line AB makes an angle α with the direction of current. Let the man travel with velocity u and at angle β to the line AB. Find time 't' by which he will be back.

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Solution

The correct option is **D**

t=S(2u cosβ)u2 cos2β−v2cos2α

In this problem, we choose axis along AB and normal to it. Since the man moves along AB, the vertical velocity components of current and man must cancel out, i.e.,u sinβ=v sinα. When the man moves from A to B , his resultant velocity along AB;=(u cosβ+v cosα)

Hence,S = (u cosβ+v cosα)t1

While for motion from B to A S = (u cosβ - v cos α)t2

From the condition of the problem,t1+t2=t

Su cosβ+v cosα+Su cosβ−v cosα

= t;S[u cosβ−v cosα+u cosβ+v cosαu2 cos2β−v2 cos2α]=t

= S(2u cosβ)u2 cos2β−v2 cos2α=t

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