A particle is moving eastwards with a velocity `5ms^(-1)`, changes its direction northwards in `10` seconds and moves with same magnitude of velocity. The average acceleration is

Options

A. `1/(sqrt(2))ms^(-2)` toward North-East
B. `1/2 ms^(-2)` towards North
C. zero
D. `1/(sqrt(2))ms^(-2)` towards North -West

Answer

D

Explanation

Given `v_(1)=+5hati` and `v_(2)=5hatj`

From above diagram
`Deltav=v_(2)-v_(1)=5hatj-5hati`
`implies|Deltav|=sqrt(2)`
`a=(|Deltav|)/t=(5sqrt(2))/10=1/(sqrt(2))ms^(-2)`
For direction
`:.tan alpha=-5/5=-1`
Average acceleration is `1/(sqrt(2))ms^(-2)`
towards north -west.