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 $45^{o}$to the vertical. What is the speed of the rain?

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

Answer 1

Answer: (A)

$5\sqrt{2}ms^{- 1}$

Answer 2

Assume north to be $\widehat{i}$directions and vertically upwards to be $\widehat{j}$directions

Let the velocity of rain be

$\overrightarrow{v} = a\widehat{i} + b\widehat{j}$

In the first case,${\overrightarrow{v}}_{g}$= velocity of girl = $5\widehat{i}$

$\therefore{\overset{\rightarrow}{v}}_{gr} = {\overset{\rightarrow}{v}}_{r} - {\overset{\rightarrow}{v}}_{g} = (a\widehat{i} + b\widehat{j}) - 5\widehat{i} = (a - 5)\widehat{i} + b\widehat{j}$

Since rain appears to fall vertically downwards.

$\therefore$a – 5 = 0 or a = 5

In the second case

${\overrightarrow{v}}_{g} = 10\widehat{i}$

$\therefore{\overset{\rightarrow}{v}}_{gr} = {\overset{\rightarrow}{v}}_{r} - 10\widehat{i} = (a\widehat{i} + b\widehat{j}) - 10\widehat{i} = (a - 10)\widehat{i} + b\widehat{j}$

$= - 5\widehat{i} + b\widehat{j}$

Since rain appears to fall at $45{^\circ}$with the vertical

$\therefore$b = -5

$\therefore$The velocity of the rain is${\overset{\rightarrow}{v}}_{r} = 5\widehat{i} - 5\widehat{j}$

Speed of the rain is

$\left| {\overset{\rightarrow}{v}}_{r} \right| = \sqrt{(5)^{2} + ( - 5)^{2}} = 5\sqrt{2}ms^{- 1}$