Question

A swimmer wishes to cross a 500 m wide river flowing at 5 km/h. His speed with respect to the water is 3 km/Now The man has to reach the other shore at the point directly opposite to his starting point. If he reaches the other shore somewhere else, he has to walk down to this point. Find the minimum distance that he has to walk. Concept of Physics - 1 , HC VERMA , Chapter "Rest and Motion : Kinematics

Answer 1

Solution :
Velocity of man vm=3km/h
BD= Horizontal distance for resultant velcoity R
X= component of resultant R=5+3cosθ
t=0.5/3sinθ
Which is same for horizontal component of velcoity 
H=BD =(5+3cos θ)(0.5 /3) sinθ
=5+3 cosθ/6 sin θ
For H to be minimum (dH/dθ)=0
d/dθ(5+3cosθ/6sinθ)=0
18(sin²θ+cos²θ)-30cosθ=0
30Cosθ=18
cosθ=-18/30=-3/5
Sinθ=√1-Cos²θ=4/5
H=5+3cosθ/6sinθ
=5+3(-3/5)/6x4/5
=25-9/24
=16/24
=2/3 km

Answer 2

2/3 is the answer for this question