Question

Suppose while sitting in a parked car, you notice a jogger approaching towards you in the side view mirror of $R=2m$. If the jogger is running at a speed of $5{ms}^{-1}$, how fast image of the jogger appear to move when the jogger is:

1. $39m$

2. $29m$

3. $19m$

4. $9m$

Answer 1

Since $R=2m,f=\frac{R}{2}=m$

Given that , $\frac{du}{dt}=-5m/s$

we have, $\frac{1}{u}+\frac{1}{v}=\frac{1}{f}$

$v=\frac{fu}{u-f}$

Differentiating w.r.t t on both sides.

$\frac{dv}{dt}=\frac{\left(f\frac{du}{dt}\right)-\left(\frac{du}{dt}\right)\left(fu\right)}{(u-f{)}^2}$

$=\frac{-f^2\frac{du}{dt}}{(u-f{)}^2}$

$\frac{dv}{dt}=-\frac{f^2}{(u-f{)}^2}\frac{du}{dt}$

(A)When jogger is 39m away,

$\frac{dv}{dt}=-\frac{1^2}{(39-1{)}^2}(-5)=3.46\times {10}^{-3}m/s$

(B)When jogger is 29m away,

$\frac{dv}{dt}=-\frac{1^2}{(29-1{)}^2}(-5)=6.38\times {10}^{-3}m/s$

(C)When jogger is 19m away,

$\frac{dv}{dt}=-\frac{1^2}{(19-1{)}^2}(-5)=0.0154m/s$

(D) when 9m away,

$\frac{dv}{dt}=0.078m/s$