Question

A man can swim in still water at 15 kmph. The river flows at 10 kmph. If he wants to cross the river
along shortest path, he should swim along a direction making an angle o with the direction of the
stream. Then
[ ]
a) sin(2/3
b) sin' (90° + 2/3) c) 90° + sin(2/3)
d) zero​

Answer 1

Answer:

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Answer 2

Answer:

The angle of the direction of the man with the direction of the

stream is  $90\textdegree + sin^{-1} (\frac{2}{3} )$.  Option c) is correct.

Explanation:

• The perpendicular path is the shortest path for the swimmer.
• When we calculate the relative velocity of an object with respect to a given frame, we consider the frame to be at rest.

The velocity of the man in still water $=15\hspace{1mm} kmph$

The velocity of the river $=10 \hspace{1mm} kmph$

If the man has to cross the river along the shortest path, he should start swimming along a direction making an angle $\alpha$ with the vertical along the west direction.

So, the angle of the direction of the man with the direction of the

stream is:

$\alpha =90\textdegree + sin^{-1} (\frac{river's\hspace{1mm} velocity }{man's \hspace{1mm} velocity})$

$\alpha =90\textdegree + sin^{-1} (\frac{10\hspace{1mm} kmph}{15\hspace{1mm} kmph} )$

$\alpha =90\textdegree + sin^{-1} (\frac{2}{3} )$

So, the angle of the direction of the man with the direction of the

stream is  $90\textdegree + sin^{-1} (\frac{2}{3} )$.

Option c) is correct.