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  4. A particle moving with a velocity equal to 0.4 m / s is subjected to an acceleration of 0.15 m / s 2 for 2 s in a direction at right angles to its direction of motion. The resultant velocity is

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Q. A particle moving with a velocity equal to $0.4\, m / s$ is subjected to an acceleration of $0.15 \,m / s ^{2}$ for $2\, s$ in a direction at right angles to its direction of motion. The resultant velocity is

Uttarkhand PMTUttarkhand PMT 2011

Solution:

There will be no change in the velocity along horizontal direction, so
$v_{x}=0.4 \,m / s .$
Velocity in perpendicular direction
$v_{y}=0+0.15 \times 2$
$=0.3 \,m / s$
So, resultant velocity,
$v=\sqrt{v_{x}^{2}+y_{y}^{2}}$
$=\sqrt{(0.4)^{2}+(0.3)^{2}} $
$=0.5 \,m / s$