Question

A swimmer swims across the river which is flowing in east direction with velocity of 3 kmh^-1. If relative velocity of swimmer with respect to water is 4 kmh^-1 in north direction, then:

(a) What is relative velocity of swimmer with respect to river bank?

(b) If river be of 5 km width, then what will be time of crossing the river?

(c) On reaching the opposite bank, what will be distance of reaching point from starting point?

Answer 1

As \(\vec v_r\) = 3 km^-1 in east; \(\vec v_m\) = 4 km h^-1 in north, d = 5 km.

(a) Velocity of swimmer with respect to river bank is \(\vec v_m\). From the figure:

If \(\vec v_m\) makes an angle θ with \(\overrightarrow{ v_rm}\), then

\(\sin\theta = \frac{|-\vec {v_r}|}{\vec{v_m}} = \frac{3kmh^{-1}}{5kmh^{-1}} = \frac{3}{5} = 0.6\ or\ \sin\theta = 0.6\)

∴ θ = sin^-1(0.6) = 37°

(b) Time taken in crossing the river

\(t = \frac{d}{v_{rm}} = \frac{5\ km}{4\ km.hr_r^{-1}} = 1.25\ hr\) 

(c) The distance which travelled along the bank of river

BC = vr x t

= 3 km.hr^-1 x 1.25 hr

= 3.75 km

= 3.75 x 1000 m

= 3750 m

Based on Umbrella Problem