Question

A particle moving, eastwards with a velocity of $5m/s$. In $10s$, its velocity changes to $5m/s$ northwards. The average acceleration in this time is

• A. Zero
• B. $\frac{1}{\sqrt{2}}m/s^2$ towards north-west
• C. $\frac{1}{\sqrt{2}}m/s^2$ towards north-east
• D. $\frac{1}{2}m/s^2$ towards north-west

Answer 1

The correct option is B $\frac{1}{\sqrt{2}}m/s^2$ towards north-west

Initial velocity, $u=5m/s$ eastwards

Final velocity, $v=5m/s$ northwards

Now, change in the velocity $|△v|=\sqrt{v^2+u^2}=\sqrt{5^2+5^2}=5\sqrt{2}$

And the direction of $△v$ is given by

$tan\theta =\frac{v}{u}=\frac{-5}{5}=-1$ (taking east direction as positive)

$\theta =-\frac{\pi }{4}$
Now, acceleration $a=\frac{△v}{t}=\frac{5\sqrt{2}}{10}=\frac{1}{\sqrt{2}}m/s^2$ in north-west direction.