A scooter going due east at 10 ms−1 turns right through an angle of

Question

A scooter going due east at 10 m

s^{- 1}

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

Options:

A . 20.0 ms−1 south eastern direction

B . Zero

C . 10.0 ms−1 in southern direction

D . 14.14 ms−1 in south-west direction

Answer: Option D
:
D
If the magnitude of vector remains same, only direction change by

θ

then

⃗ △ v

{⃗ v}_{2}

{⃗ v}_{1}

⃗ △ v

{⃗ v}_{2}

+ (-

{⃗ v}_{1}

)
Magnitude of change in vector |

⃗ △ v

| = 2 v sin (

\frac{θ}{2}

)
|

⃗ △ v

| = 2

\times

\times

sin(

\frac{90^{\circ}}{2}

) = 10

\sqrt{2}

= 14.14 m/s
Direction is south - west as shown in figure.

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