1
You visited us 1 times! Enjoying our articles? Unlock Full Access!
Question

# A man wants to swim across a river from A to B and back from B to A always following line AB. The distance S between points A and B is S. The velocity of the river current v is constant over the entire width of the river. The line AB makes an angle α with the direction of current. Let the man travel with velocity u and at angle β to the line AB. Find time 't' by which he will be back.

A

t=S(2u cosβ)v2 cos2βu2 cos2α

No worries! We‘ve got your back. Try BYJU‘S free classes today!
B

t=S(2u sinα)u2 sin2βv2 sin2α

No worries! We‘ve got your back. Try BYJU‘S free classes today!
C

t=S(20 sinα)v2 sin2βu2 sin2α

No worries! We‘ve got your back. Try BYJU‘S free classes today!
D

t=S(2u cosβ)u2 cos2βv2cos2α

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
Open in App
Solution

## The correct option is D t=S(2u cosβ)u2 cos2β−v2cos2α In this problem, we choose axis along AB and normal to it. Since the man moves along AB, the vertical velocity components of current and man must cancel out, i.e.,u sinβ=v sinα. When the man moves from A to B , his resultant velocity along AB;=(u cosβ+v cosα) Hence,S = (u cosβ+v cosα)t1 While for motion from B to A S = (u cosβ - v cos α)t2 From the condition of the problem,t1+t2=t Su cosβ+v cosα+Su cosβ−v cosα = t;S[u cosβ−v cosα+u cosβ+v cosαu2 cos2β−v2 cos2α]=t = S(2u cosβ)u2 cos2β−v2 cos2α=t

Suggest Corrections
0
Join BYJU'S Learning Program
Related Videos
Relative Motion in 2D
PHYSICS
Watch in App
Explore more
Join BYJU'S Learning Program