One of our academic counsellors will contact you within 1 working day.

Please check your email for login details.
MY CART (5)

Use Coupon: CART20 and get 20% off on all online Study Material

ITEM

DETAILS

MRP

DISCOUNT

FINAL PRICE

Total Price: Rs.

There are no items in this cart.

Continue Shopping

Continue Shopping

Menu

## A river is flowing from west to east at speed of 5 m/minute . A man on south bank of river, capable of swimming at 10 m/min in still water, wants to swim across river in shortest time. He should swim in direction a)due north. b)30 east of north. c) 30 north of west. d) 60 east of north. Justify you answer.

11 years ago

-------------> V river = 5m/min

V man = 10 m/minsuppose width of river is x mts

Case 1if he travels due north

then his resultant velocty ( due to velocity of downstream of water) acts 30 north of east and the distane to be covered to reach the other bank increses to = x/cos 30

i.e., 2x /3

^{1/2}and resultant velocity is 11.2 m/minso time taken is

x / 9.7mins

Case 2if he travels due 30 north of east

then his resultant velocty ( due to velocity of downstream of water) actsdue north and the distane to be covered to reach the other bank will be x mts and resultant velocity is 8.7 m/min

so time taken is

x / 8.7minsSo case 1 takes least time

he should travel due north(a)

11 years ago

he should swim due north!!!

his velocity in moving water is equal to his own velocity plus the velocity of the river.

now to cross the river in smallest possible time he should maximise the velocity in the north direction.

suppose he move at an angle of @ from the bank then the component due north is 10sin@m/min. because river velocity is zero in the north direction.

now this velocity due noth will be maximum when @ tends to 90 degrees.

hence he should move due north that is 90 degrees to the bank.

11 years ago

Thus ,the resultant of the Velocity of man in still water (Vm)(=OA)and the Velocity of water (Vw)(=OB) is along OP, and is given by (V)(=OC)In right triangle OCA , we haveSin x= AC/OA=Vw/Vm =5/10 = 1/2x=30°Therefore, the man should start swimming at an angle of 30° (west of north)

3 years ago

Yes we should move due north to cover the distance in the shortest time as then the velocity of river and man will get added up and distance would be covered in the shortest time

2 years ago

If the man crossed the river with a angle @ to examined the river velocity then he should travel due north with velocity vcos@.so time taken to cross the river is d/vcos@.after differentiate both the sides of eqn we get @=0. So he should travel in north direction.

one year ago

hello student

The given condition can be shown as infigure

Time taken to cross the river,t=d/Vscosθ

for time to be minimum,cosθ=maximum

⇒θ=0

So, summer should swim due North.

hope it helps

thankyou

5 months ago

-------------> V river = 5m/min

V man = 10 m/minsuppose width of river is x mts

Case 1if he travels due north

then his resultant velocty ( due to velocity of downstream of water) acts 30 north of east and the distane to be covered to reach the other bank increses to = x/cos 30

i.e., 2x /3

^{1/2}and resultant velocity is 11.2 m/minso time taken is

x / 9.7mins

Case 2if he travels due 30 north of east

then his resultant velocty ( due to velocity of downstream of water) actsdue north and the distane to be covered to reach the other bank will be x mts and resultant velocity is 8.7 m/min

so time taken is

x / 8.7minsSo case 1 takes least time

HENCE HE SHOULD TRAVEL TOWARDS NORTH.

SO A OPTION IS CORRECT

4 months ago

Dear Student

In order to swim across the river in the shortest time, the man should swim straight due north.

Because the velocity of the river is west to east and there is no component in north-south. So, it will not affect the man's time in order to swim to the other bank.

I hope this answer will help you.

Thanks & Regards

Yash Chourasiya

4 months ago

Think You Can Provide A **Better Answer ?**

Answer & Earn Cool Goodies

Have any Question? Ask Experts

Post Question

copyright © 2006-2020 askIITians.com

info@askiitians.com