Question

A man crosses a river in a boat. If he cross the river in minimum time he takes 10 minutes with a drift 120 m. if he crosses the river taking shortest path. He takes 12.5 minute find

(a) width of the river

(b) velocity of the boat with respect to water

(c) speed of the current

Answer 1

☜☆☞ River Man Problem : Kinematics :

Final Answer : (a) 200 m
(b) 1/3 m/s
(c) 0.2 m/s

Steps:
1) We are assuming respective velocities are same in both cases and u > v. (see below)

Case 1 : (Crossing in Minimum Time)
=> When man wants to cross the river in minimum time, then he must swim perpendicular to flow of river.

Let the relative velocity of man wrt river be 'v' &
Velocity of river /Speed of current be 'u'.
d - Width of River

2) Time taken to cross the river in minimum time :$t_1 = 10 *60 =600 s$

Drift = u * t
=> 120 m = u * 600
=> u = 0.2m/s
Therefore,
(c) Speed of current = 0.2m/s

(3) We can also, say :
$t_1 = \frac{d}{v} \\ => d = 600 v --(1)$

4) Case - 2 (Crossing with shortest Path) :
Let the relative velocity 'v' make an angle $\theta$ with vertical in this case .

$t_2 = 12.5*60 s .$
For Drift to be minimum, (as v < u)
Horizontal component of 'v' must equal to speed of river.
That is,
$v \sin(\theta) = u \\ => \sin(\theta) = \frac{u}{v} \\ => \cos(\theta) =\sqrt{1- \frac{u^2}{v^2}}--(2)$ .

5) Now,
We can say,
$t_2 = \frac{d}{v \cos(\theta)} \\ => 12.5 * 60 = \frac{600v }{v \cos(\theta) } \\ => 1.25^2 = \frac{v^2}{v^2-u^2} \\ => v = \frac{u*1.25}{\sqrt{1.25^2-1}} \\ => v = \frac{1}{3} m/s \:\:( \:as\: u =0.2m/s)$

6) Now,
$d = 600*v =200 m.$

^_° Hope You got your desired answer.

Answer 2

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