Question

A river flows 3 kmh^-1 and a man is capable of swimming 2 km h^-1. He wishes to cross it in minimum time. In which direction will he swim?

• A. (a) sin^-1(2/3)
• A. (b) cos^-1(2/3)
• A. (c) tan^-1(2/3)
• A. (d) cot^-1(2/3)

Answer 1

(a)

Let us assume he swims at an angle θ with the perpendicular as shown. If river is lm wide time taken to cross it

t = l/2cos θ, v_x = 3 – 2 sin θ horizontal distance covered along x direction during this period x = v_x.

t = (3 – 2 sin θ) l/2cosθ

for t to be min dx/dθ = 0, or

l[3/2 secθ tan θ - sec²θ] = 0 or

sin θ = 2/3