A scooter going due east at 10, ms^{-1} turns right through an angle of 90^° . If speed of scoot

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3-2.Motion in Plane

hard

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

A$20.0\, ms^{-1}$ south eastern direction

BZero

C$10.0\, ms^{-1}$ in southern direction

D$14.14\, ms^{-1}$ in south-west direction

Solution

(d) If the magnitude of vector remains same, only direction change by $\theta $ then
$\overrightarrow {\Delta v} = \overrightarrow {{v_2}} – \overrightarrow {{v_1}} $, $\overrightarrow {\Delta v} = \overrightarrow {{v_2}} + ( – \overrightarrow {{v_1})} $
Magnitude of change in vector $|\overrightarrow {\Delta v} |\, = 2v\sin \left( {\frac{\theta }{2}} \right)$
$|\overrightarrow {\Delta v} |\, = 2 \times 10 \times \sin \left( {\frac{{90^\circ }}{2}} \right)$= $10\sqrt 2 $= $14.14\,m/s$
Direction is south-west as shown in figure.

Standard 11

Physics

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