Question

$\huge{\red{\underline{\underline{Question}}}}$

Two boats A and B move away from buoy anchored in the middle of the river along mutully perpendicular straight lines, the boat A moves along the river and boat B across the river.

Having moved off an equal distance fronm the buoy the boat returned.

Find the times of motion of boats $\sf{\dfrac{t_A}{t_B}}$ if the velocity of each boat with respect to water is
n = 1.2 times greater than the stream velocity.​

Answer 1

♊♊YOUR ANSWERS♍♍

✍tA/tB = 1.8

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EXPLANATION —

↪Actually welcome to the concept of Relative Velocity

Let l be the distance covered by the boat A along the river as well as by the boat B across the river.

Let v0 be the stream velocity and v' the velocity of each boat with respect water. Therefore time taken by the boat A in its journey

$tA = \frac{l}{v' + v0} + \frac{l}{v' - v0}$

and for the boat B

$tB = \frac{l}{ \sqrt{ {v'}^{2} + {v0}^{2} } } + \frac{l}{ \sqrt{ {v'}^{2} - {v0}^{2} } }$

Therefore

$\frac{tA}{tB} = \frac{v'}{ \sqrt{ {v'}^{2} + {v0}^{2} } }$

$\frac{tA}{tB} = \frac{n}{ \sqrt{ {n}^{2} - 1} } \: \: \: n = \frac{v'}{v0}$

By substituting the values

$\frac{tA}{tB} = 1.8$

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