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  • Solve! A man wishes to cross a river in a boat. If he the river in minimum time he takes 10 minutes with a drift of 120m. If he crosses the river taking the shortest route it takes 12.5 min. The velocity of boat with repect to water is

A man wishes to cross a river in a boat. If he  the river in minimum time he takes 10 minutes with a drift of 120m. If he crosses the river taking the shortest route it takes 12.5 min. The velocity of boat with repect to water is 

  • Option 1) (1/3) m/s

     

  • Option 2) (1/4) m/s

     

  • Option 3) (1/5) m/s

     

  • Option 4) (1/6) m/s

     

 

Answers (1)

As we have learnt,

 

Boat - River Problem -

To cross river in the shortest path

t= \frac{d}{\sqrt{v^{2}-u^{2}}}


 

- wherein

d= width of river 

v    = Speed of Boat w.r.t. river

u    = speed of river

 

 Let Velocity of man is v and velocity of river is u.

For minimum time

t_1 = \frac{d}{v} = 10\;min\;\;-(1)

drift = u\cdot t_1 = 120m

or u\cdot \frac{d}{v} = 120 \Rightarrow u = \frac{120}{600} m/s \Rightarrow u =\frac{1}{5} m/s

For Shortest route

Let it makes angle \theta with river flow.

\\*\Rightarrow v\cos\theta = u \\* \& \;\;t_2 = \frac{d}{v\sin\theta} = \frac{d}{v\cdot\sqrt{1-\frac{u^2}{v^2}}} = \frac{d}{\sqrt{v^2 - u^2}} \\* or \; \frac{d}{\sqrt{v^2 - u^2}} = 12.5 min \;\;-(2)

DIvide (2) in (1)

\\*\Rightarrow \frac{d}{v\cdot \frac{d}{\sqrt{v^2- u^2}}} = \frac{10}{12.5} \; or \;\sqrt{1-\frac{u^2}{v^2}} = \frac{4}{5} \\* 1-\frac{u^2}{v^2} = \frac{16}{25} \;or \;\frac{u^2}{v^2} = \frac{9}{25} \\* \Rightarrow \frac{u}{v} = \frac{3}{5} \Rightarrow v = \frac{5}{3}\cdot u = \frac{5}{3}\times \frac{1}{5} = \frac{1}{3}m/s

    


Option 1)

\frac{1}{3}m/s

Option 2)

\frac{1}{4}m/s

Option 3)

\frac{1}{5}m/s

Option 4)

\frac{1}{6}m/s

Posted by

subam

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