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Rain is falling vertically with a speed of $12 \mathrm{~ms}^{-1}$. A woman rides a bicycles with a speed of $12 \mathrm{~ms}^{-1}$ in east to west direction. What is the direction in which she should hold her umbrella?
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Verified Answer
The correct answer is:
$45^{\circ}$, towards east
The given situation is shown below
Here, velocity of rain, $v_{r}=12 \mathrm{~ms}^{-1}$
and velocity of woman, $v_{w}=12 \mathrm{~ms}^{-1}$
The relative velocity of rain w.r.t. woman is
$\begin{aligned}
v_{r w} &=\sqrt{v_{w}^{2}+v_{r}^{2}} \\
&=\sqrt{(12)^{2}+(12)^{2}}=12 \sqrt{2} \mathrm{~ms}^{-1}
\end{aligned}$
The direction in which the woman should hold her umbrella is
$\begin{aligned}
\sin \theta &=\frac{v_{w}}{v_{r w}}=\frac{12}{12 \sqrt{2}} \\
\Rightarrow \quad \theta &=\sin ^{-1}\left(\frac{1}{\sqrt{2}}\right)=45^{\circ}, \text { towards east }
\end{aligned}$
Here, velocity of rain, $v_{r}=12 \mathrm{~ms}^{-1}$
and velocity of woman, $v_{w}=12 \mathrm{~ms}^{-1}$
The relative velocity of rain w.r.t. woman is
$\begin{aligned}
v_{r w} &=\sqrt{v_{w}^{2}+v_{r}^{2}} \\
&=\sqrt{(12)^{2}+(12)^{2}}=12 \sqrt{2} \mathrm{~ms}^{-1}
\end{aligned}$
The direction in which the woman should hold her umbrella is
$\begin{aligned}
\sin \theta &=\frac{v_{w}}{v_{r w}}=\frac{12}{12 \sqrt{2}} \\
\Rightarrow \quad \theta &=\sin ^{-1}\left(\frac{1}{\sqrt{2}}\right)=45^{\circ}, \text { towards east }
\end{aligned}$
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