A particle slides with a speed of `3m s^(-1)` at P. When it reaches `Q`,it acquires a speed of `4m s^(-1)` after describing an angle of `60^(@)` at `O` as shown in figure .Find the changes in the velocity of the particle between `P` and `Q`. Assume that the path followed by the particle is circular from `P` to `Q`

Options

A. `sqrt(10m)s^(-1)`
B. `sqrt(9m)s^(-1)`
C. `sqrt(13m)s^(-1)`
D. `sqrt(15m)s^(-1)`

Answer

C

Explanation

As the particle moves from P to Q, its speed increases from `v_(1)=3m s^(-1)` to `v_(2)=4m s^(-1)` and the angle described by the velocity vector is equal to `theta=60^(@)`.Then the change in velocity is given as
`Deltav=sqrt(v_(1)^(2)+v_(2)^(2)-2v_(1)v_(2)cos theta)`
Substituting, `v_(1)=3m s^(-1),v_(2)=4m s^(-1) ` and `theta=60^(@)` in the above expression ,we have
`Deltav=sqrt(3^(2)+4^(2)-2.3.4cos60^(@))`
This yield `Deltav=sqrt(13m)s^(-1)` and the change in velocity is directed as shown in figure
.