Question

During a rainy day, rain is falling vertically with a velocity $2m{s}^{-1}$. A boy at rest starts his motion with a constant acceleration of $2m{s}^{-2}$ along a straight road. Find the rate at which the angle of the axis of umbrella with vertical should be changed so that the rain always falls parallel to the axis of the umbrella.

Answer 1

Velocity of rain with respect to boy

${\overrightarrow{v}}_{rain.boy}={\overrightarrow{v}}_{rain}-{\overrightarrow{v}}_{boy}={\overrightarrow{v}}_{rain}+(-{\overrightarrow{v}}_{boy})$

At any time t, rain will appear to the boy as shown in Fig.

$\tan \theta =\frac{at}{v}$

Boy should hold his umbrella at an angle $\theta$ from the vertical

$\therefore \tan \theta =\frac{at}{v}$

To find the rate at which the angle of the axis of umbrella with vertical, we need to find $d\theta /dt$.

$\frac{dtan\theta }{dt}=\frac{d\left(\frac{at}{v}\right)}{dt}\Rightarrow se{c}^2\theta \frac{d\theta }{dt}=\frac{a}{v}$

$\Rightarrow \frac{d\theta }{dt}=\frac{a}{vse{c}^2\theta }=\frac{a}{v[1+ta{n}^2\theta ]}$

$=\frac{a}{v[1+\frac{a^2{t}^2}{v^2}]}=\frac{av}{v^2+a^2{t}^2}$

$=\frac{2\times 2}{4+4t^2}=\frac{1}{1+t^2}$

$\Rightarrow \frac{d\theta }{dt}=\frac{1}{1+t^2}$