Bottle falls during Boating ⌢ ¨ \ddot \frown

Classical Mechanics Level 3

A boat travels upstream in a river with a constant but unknown speed v v with respect to water. At the start of this trip upstream, a bottle is dropped over the side. After 7.5 m i n u t e s 7.5~minutes the boat turns around and heads downstream. It catches up the bottle when the bottle has drifted 1 k m 1~km downstream from the point at which it was dropped into the water. Then the velocity of water current is (in k m / h r km/hr )

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Tanishq Varshney by Tanishq Varshney , 23, India

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1 solution

Let V be the boat velocity, U that of river. So the total time for whole operation is 1 k m U k m / h r = 1 U h r . Time for boat to move up is 1/8 hr. It moved (V -U) * 1/8 km up. So time to move down stream (1/U - 1/8)hr. ∣ t e x t T h i s i s a t ( V + U ) k m / h r . Distance moved down stream is ..1 + (V -U) * 1/8 km at (V+U) km/hr and took (1/U - 1/8)hr. V e l . = D i s . T i m e . ∴ ( V + U ) = 1 + ( V − U ) ∗ 1 / 8 1 / U − 1 / 8 (V+U) * (1/U - 1/8 )= 1 + (V -U) * 1/8 . S o l v i n g , U = 4 k m / h r . \text{Let V be the boat velocity, U that of river. } \\\text{So the total time for whole operation is } \dfrac{1 km}{U km/hr}=\dfrac1U hr.\\ \text{Time for boat to move up is 1/8 hr.}\\\text{ It moved (V -U) * 1/8 km up. }\\\text{ So time to move down stream (1/U - 1/8)hr.}\\|text{This is at (V+U) km/hr .}\\\text{Distance moved down stream is ..1 + (V -U) * 1/8 km} \\\text{ at (V+U) km/hr and took (1/U - 1/8)hr.}\\ Vel. =\dfrac{Dis.}{Time.}~~\therefore(V+U)=\dfrac{1 + (V -U) * 1/8 }{1/U - 1/8 }\\\text{ (V+U) * (1/U - 1/8 )= 1 + (V -U) * 1/8 . }\\ Solving, ~~U =4 km/hr.

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Correct minor LaTeX Bugs in the solution. And its good to use lowercase letters * for Variables like velocity, distance, time. Because uppercase is not recommended in LaTeX.

Thank you for the nice solution.

Muhammad Arifur Rahman - 5 years, 8 months ago

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