Question

A man wishes to cross a river in a boat. If he crosses the river in minimum time he takes 10 minutes with a drift of 120 m. If he crosses the river taking the shortest route he takes 12.5 minutes. The magnitude of velocity of the boat w.Ct. Water is

Answer 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..

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Answer 2

Answer:

Explanation:

Width = d

Man = V

River = v

In minimum time, he goes perpendicular to the river.

d/V = 10 min------(1)

v x 10 min = 120 m, so v = 12 m /min

In minimum distance, the resultant of the velocities is perpendicular to river

If he is at angle x with the perpendicular

V sinx = v = 12-----(2)

V cosx = d / 12.5-----(3)

Squaring and adding (2) and (3)

2V^2 = 144 + d^2 / 156.25 -----(4)

Putting value of V from (1) in (4),

d^2 /50 = 144 + d^2 /156.25

d^2 = 10500 m

d = 100m

There could be some calculation error, but the method is correct.

Explanation: