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Question

# A man wants to reach point B on the opposite bank of a river flowing at a speed as shown in figure. What minimum speed relative to water should the man have so that he can reach point B?

A
2u
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B
u2
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C
3u
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D
u3
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Solution

## The correct option is B u√2Given, Speed of river =u Assuming, Speed of man swimming =v in direction 'θ' anticlock with line AC On breaking speed of man v into x and y component vx=vsinθ vy=vcosθ Let us suppose vn is the net speed of the man, we know that to reach point B from A, vn should be along AB. Also, x component of vn, vnx=u−vsinθ y component of vn, vny=vcosθ In ΔACB, tan45∘=vnxvny ⇒1=u−vsinθvcosθ ⇒vcosθ=u−vsinθ ⇒vcosθ+vsinθ=u ⇒v(cosθ+sinθ)=u ⇒v=ucosθ+sinθ...(1) Here, to get minimum value of v, value of denominator should be maximum. Assuming, t=cosθ+sinθ ∴dtdθ=−sinθ+cosθ [Applying concept of maxima and minima] For maxima or minima, dtdθ=0⇒−sinθ+cosθ=0 ⇒cosθ=sinθ ⇒1=tanθ ⇒θ=45∘ Now, d2tdθ2=−cosθ−sinθ =−(cosθ+sinθ) =−(cos45∘+sin45∘)=−√2 = Negative value d2tdθ2<0 ∴t has maximum value at θ=45∘ On putting θ=45∘ in equation (1) vmin=ucos45∘+sin45∘=u√2 vmin=u√2

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