A scooter going due east at 10 ms –1 turns right through an angle of 90°. If the speed of the scooter remains unchanged in taking turn, the change is the velocity of the scooter is
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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 + (-\vec{v}_1)
Magnitude of change in vector \(\left|\overrightarrow{\Delta v}\right| = 2v \sin\left(\frac{\theta}{2}\right)\)
\(\left|\overrightarrow{\Delta v}\right| = 2 \times 10 \times \sin\left(\frac{90^\circ}{2}\right)\) = \(10\sqrt{2}\) = 14.14 m/s
Direction is south-west as shown in figure.
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