Question

A girl riding a bicycle with a speed of $5\,ms^{-1}$ towards north direction, observes rain falling vertically down. If she increases her speed to $10\,ms^{-1}$, rain appears to meet her at $45^°$ to the vertical. What is the speed of the rain?

• A. $5\sqrt{2}\,ms^{-1}$
• B. $5\,ms^{-1}$
• C. $10\sqrt{2}\,ms^{-1}$
• D. $10\,ms^{-1}$

Answer 1

Answer: (A)

Assume north to be $\hat{i}$ direction and vertically upwards to be $\hat{j}$ direction.

Let the velocity of rain be $\vec{v_r}=a\hat{i}+b\hat{j}$
In the first case, $\vec{v}_{g}=$ velocity of girl $=5\hat{i}$
$\therefore \vec{v}_{rg}=\vec{v_{r}}-\vec{v}_{g}$
$=\left(a\hat{i}+b\hat{i}\right)-5\hat{i}$
$= \left(a - 5\right)\hat{i}+b\hat{j}$
Since rain appears to fall vertically downwards,
$\therefore a-5=0$ or
$a=5$
In the second case,
$\vec{v}_{g}=10\hat{i}$
$\therefore \vec{v}_{rg}=\vec{v}_{r}-10\hat{i}$
$=\left(a\hat{i}+b\hat{j}\right)-10\hat{i}$
$=\left(a-10\right)\hat{i}+b\hat{j}$
$=-5\hat{i}+b\hat{j}$
Since rain appears to fall at $45^{°}$ with the vertical,
$\therefore a = b =-5$
$\therefore $ The velocity of the rain is
$\vec{v}_{r}=5\hat{i}-5\hat{j}$
Speed of the rain is
$\left|\vec{v}_{r}\right|=\sqrt{\left(5\right)^{2}+\left(-5\right)^{2}}$
$=5\sqrt{2}\,ms^{-1}$