Q.
A fish rising vertically up towards the surface of water with speed $3\ ms^{- 1}$ observes a bird diving vertically down towards it with speed $9\ ms^{- 1}$. The actual velocity of bird is
A) 4.5 $ms^{- 1}$
B) 5. $ms^{- 1}$
C) 3.0 $ms^{- 1}$
D) 3.4 $ms^{- 1}$
Explanation
(a) Here optical distance between fish and the bird is $s = y^{'} + \mu y$
Differentiating $w.r.t.$ we get $\frac{ds}{dt} = \frac{dy^{'}}{dt} + \frac{\mu dy}{dt}$
$\Rightarrow 9 = 3 + \frac{4}{3}\frac{dy}{dt} \Rightarrow \frac{dy}{dt} = 4.5\ m/s$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.