Question

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

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

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 $\overrightarrow{v_r}=a\hat{i}+b\hat{j}$
In the first case, ${\vec{v}}_g=$ velocity of girl $=5\hat{i}$
$\therefore {\vec{v}}_{rg}=\overrightarrow{v_r}-{\vec{v}}_g$
$=\left(a\hat{i}+b\hat{i}\right)-5\hat{i}$
$=\left(a-5\right)\hat{i}+b\hat{j}$
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
$\left|{\vec{v}}_r\right|=\sqrt{{\left(5\right)}^2+{\left(-5\right)}^2}$
$=5\sqrt{2}m{s}^{-1}$