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A scooter going due east at $10 \mathrm{~m} / \mathrm{s}$ turns right through an angle of $90^{\circ}$. If the speed of the scooter remains unchanged in taking turn, the change is the velocity of the scooter is
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$14.14 \mathrm{~m} / \mathrm{s}$ in south-west direction.
If the magnitude of vector remains same, only direction change by $\theta$ then
$\vec{\Delta v}=\vec{v_2}-\vec{v_1}, \vec{\Delta v}=\vec{v_2}+\left(-\vec{v_1}\right)$
Magnitude of change in vector $|\overrightarrow{\Delta v}|=2 v \sin \left(\frac{\theta}{2}\right)$
$|\overrightarrow{\Delta v}|=2 \times 10 \times \sin \left(\frac{90^{\circ}}{2}\right)=10 \sqrt{2}=14.14 \mathrm{~m} / \mathrm{s}$
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