A swimmer can swim with velocity 12 km/h in still water. Water flowing in a river has velocity 6 km/h. The direction with respect to the direction of flow of river water he should swim in order to reach the point on the other bank just opposite to his starting point is __________ (in degrees).
(Round off to the nearest integer)
Correct answer is '120'. Can you explain this answer?

Question

A swimmer can swim with velocity 12 km/h in still water. Water flowing in a river has velocity 6 km/h. The direction with respect to the direction of flow of river water he should swim in order to reach the point on the other bank just opposite to his starting point is __________ (in degrees).
(Round off to the nearest integer)

Correct answer is '120'. Can you explain this answer?

Answer

Explanation:

Let the width of the river be d km.

Let the swimmer swim at an angle of θ with respect to the direction of flow of river water.

Velocity of swimmer in still water = 12 km/h.

Velocity of river water = 6 km/h.

Velocity of swimmer with respect to river water = 12 cosθ - 6 km/h.

Time taken to cross the river = d / (12 cosθ - 6) hours.

Distance travelled downstream in this time = 6(d / (12 cosθ - 6)) km.

Distance travelled upstream in this time = 6(d / (12 cosθ - 6)) km.

Total distance travelled = √((6(d / (12 cosθ - 6)))^2 + d^2) km.

Time taken = √((6(d / (12 cosθ - 6)))^2 + d^2) / (12 cosθ - 6) hours.

The swimmer wants to reach the point on the other bank just opposite to his starting point.

So, the total distance travelled should be 2d km.

Therefore, 2d = √((6(d / (12 cosθ - 6)))^2 + d^2).

Solving this equation, we get cosθ = 1/2.

So, θ = 60°.

But the swimmer has to swim at an angle of 60° with respect to the direction of flow of river water.

So, the angle with respect to the direction of the bank is 180° - 60° = 120°.

Hence, the swimmer should swim at an angle of 120° with respect to the direction of flow of river water.

Answer

v_m = 12 km/h
v_r = 6 km/h
vm sin θ = vr
sin θ = 6/12
θ = 30°
α = 90 + θ
α = 90 + 30
α = 120°

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