Question

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

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

Answer 1

The correct option is A 5 $\sqrt{2}$ $m{s}^{-1}$

Let us assume north is in $\hat{i}$ direction and vertically downwards is $-\hat{j}$ direction

Let the velocity of rain $\overrightarrow{v_r}=a\hat{i}+b\hat{j}$

Given, Velocity of girl, $\overrightarrow{v_g}=5\hat{i}$

Velocity of rain w.r.t girl, $\overrightarrow{v_{rg}}=(a-5)\hat{i}+b\hat{j}$

According to question, rain appears to fall downwards

Hence, $a-5=0⟹a=5$

Given, velocity of girl increased to , $\overrightarrow{v_g}=10\hat{i}m/s$

Velocity of rain appears to fall to the girl, $\overrightarrow{v_{rg}}=\overrightarrow{v_r}-\overrightarrow{v_g}=(a-10)\hat{i}+b\hat{j}$

Now the rain appears to fall at ${45}^o$

Hence , $b=a-10⟹b=-5$

$\therefore$ Velocity of rain, $\overrightarrow{v_r}=5\hat{i}-5hatj$

$\therefore$ Speed of rain, $|v_r|=\sqrt{5^2+5^2}=5\sqrt{2}m/s$