Explanation
The speed of the man is 2m/s2m/s and the speed of the water current is √3m/s√3m/s.
The velocity component acting opposite to that of the velocity of water current is given as,
vmsinθ=vrvmsinθ=vr
2×sinθ=√32×sinθ=√3
θ=60∘θ=60∘
The angle between the speed of man and the water current is given as,
θ1=60∘+90∘θ1=60∘+90∘
=150∘=150∘
Thus, the direction of swimming is at an angle 150∘150∘ to the water current.
Let velocity of water be u and velocity of boat in still water be v
Let distance travelled be d
We need to find d/v
t1= d/(v+u)
d= t1 v + t1 u (1)
t2= d/(v-u)
d = t2 v- t2 u (2)
From 1&2
v t1 + u t1= v t2- u t2
u(t1+ t2) = v(t2- t1)
u = v(t2-t1) / (t1+t2)
Substituting value of u in (1)
d = t1v + t1 [v(t2-t1) /t1+t2)] t1
d(t1+t2) = t1v(t1+t2) + vt1 (t2-t1)
d(t1+t2) = t1vt1 + t2t1v + vt1t2 - t1t1v
d(t1+t2) =2vt1t2
d/v = 2t1t2/( t1+t2)
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