Question

Q] A man can swim in still water with speed v. he wants to cross the river of width d that flows with speed u and reaches directly opposite to his starting point.

(a) In which direction should he try to swim ? (i.e find the angle he makes with the river flow)

(b) How much time will he take to cross the river

#Que's for neet aspirants <(￣︶￣)>​

Answer 1

QUESTION

A man can swim in still water with speed v. he wants to cross the river of width d that flows with speed u and reaches directly opposite to his starting point.

prblem 1

In which direction should he try to swim ? (i.e find the angle he makes with the river flow)

solution 1

please refer to the attachment

problem 2

How much time will he take to cross the river

solution 2

please refer to the attachment

there are three attachment please refer to them

solved

Answer 2

☆Concept:

see the solution with refer to above diagram!

》The man will have two velicities which will affect its motion, v(man's speed) and u(river speed).

》As man will try to move straight from 'A' to 'B', the flow of river will tilt the man towards it by which it can cause a little displacement from its actual position to reach.

》So the man has to go in such a direction, that even after the tilt, he should reach the desired location.

☆Solution:

a) In which direction should he try to swim ?

• we can see in the figure, that the direction is greater than 90° from river axis, so that tilt make him go straight.

》Main point to note:

• For moving straight from A to B, the displacement in x- direction should be zero.
• That also means the velocity in x- direction should be zero.
• so the man will move in such a way, that will cancel the effect of velocity of river in x-dirn. and only component of v will be responsible for motion.

》From above points:

• $\sf vsin \alpha = u$

• $\sf sin \alpha = \dfrac{u}{v}$

• $\sf \alpha = sin{-1}( \dfrac{u}{v})$

• direction will be from +x-axis:

》$\sf \theta = 90° + \alpha$

• $\bf \theta = \dfrac{\pi}{2} + sin{-1}( \dfrac{u}{v})$

(b) How much time will he take to cross the river?

》$\sf time= \dfrac{ displacement}{velocity}$

Note: Take that component of velocity which is responsible for disp. in y‐direction.

• $\sf t = \dfrac{d}{vcos \alpha }$

☆ $\ sin² \alpha + cos² \alpha = 1$

-- $\ cos\alpha = \sqrt {1- sin² \alpha }$

-- $\ cos\alpha = \sqrt {1- \dfrac{u²}{v²} }$

-- $\ cos\alpha = \sqrt { \dfrac{ v²- u²}{ v²}}$

-- $\ cos\alpha = \dfrac{ \sqrt{ v²- u²}}{ v}$

• $\sf t = \dfrac{d.(v)}{v. \sqrt{ v²-u²}}$

• $\bf{ t=( \dfrac{d}{\sqrt{ v²- u²}})}$

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