(a) Given. Speed of river, \(v_r\) = 3 m/s (east)
Velocity of summer (w.r.t. river), \(v_s\) = 4 m/s (north)
Y – component of his resultant velocity = 4 m/s (north)
x – component (produced by river flow) = 3 m/s (east)
Resultant velocity, v = \(\sqrt{v^2_r+v^2_s}\)
= \(\sqrt{(3)^2+(4)^2}=\sqrt{3+16}\)
= \(\sqrt{25}\) = 5 m/s.
∴ tan θ = \(\frac{v_r}{v_s}=\frac{3}{4}\) = 0.75
tan 36º54’
⇒ θ = 36º54’ ≈ θ = 37º N
(b) The swimmer should swim at an angle of approx. of 37º north. (If he wants to reach on the point directly opposite to him).
Resultant speed of swimmer,
v = \(\sqrt{v^2_s-v^2_r}=\sqrt{4^2-3^2}\)
= \(\sqrt{16-9}=\sqrt7\,m/s\).
tan θ = \(\frac{v_r}{v}=\frac{3}{\sqrt7}\)
⇒ \(\theta\) = \(tan^{-1}(\frac{3}{\sqrt7})\) with north
(c) From (a),
Time taken by swimmer (to cross river), \(t_1=\frac{d}{v_s}=\frac{d}{4} s\)
From (b)
Time taken to cross river,\(t_1=\frac{d}{v}=\frac{d}{\sqrt7}\)
As, \(\frac{d}{4}<\frac{d}{\sqrt7},\,∴\,t_1<t_2\)
Hence, in case (a), the swimmer will cross the river in shorter time.
