First slide

Relative motion in 2 dimension

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?

Moderate

A

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

$5 {ms}^{- 1}$
C

$10 \sqrt{2} {ms}^{- 1}$
D

$10 {ms}^{- 1}$

Solution

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

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