Question

A particle is moving with a velocity of 10 m/s towards east.After 10s its velocity changes to 10 m/s towards north.Its average accelaration is ?

1. ZERO
2. √2 m/s^2 towards N-E
3. 1/√2 m/s^2 towards N-E
4. 1/√2 m/s^2 towards N-W

Answer 1

Average acceleration is defined as- (change in velocity/time elapsed)

Here,change in velocity = `sqrt(10^2+10^2)` or `10sqrt2 m/s`

So,average acceleration is `(10sqrt2)/10 m/s^2` i.e `sqrt2 m/s^2`

As,both the velocity vectors are equal in magnitude,their resultant will be at angle of 45 degrees w.r.t each other. And that will be the direction of the average acceleration as well,i.e along north west

Note: **Change in acceleration means,final velocity vector(`v`) - initial velocity vector`(v'`)
= `v+(-v')`

As `v'` vector is located towards east,so `-v'` will be located towards west,so a resultant of two vectors of equal magnitude one directed along north and one along west will be along north west**