A body moving in a curved path possesses a velocity 3 m/s towards north at any instant of its motion .After 10s ,the velocity of the body was found to be 4 m/s towards west.Calculate the average acceleration during this interval.

Explanation

To solve this problem the vector nature of velocity must be taken into account.
In the figure,the initial velocity `v_(0)` and the final velocity v are drawn common origin .The vector difference if them is found by the parallelogram method.
The magnitude of difference is
`|v-v_(0)|=OC=sqrt(OA^(2)+AC^(2))`
`sqrt(4^(2)+3^(2))`=5m/s
Tghe direction is given by
`tantheta=(3)/(4)=0.75.theta=37^(@)`
Average acceleration `=(|vecv-vecv_(0)|)/(t)=(5)/(10)=0.5m//s^(2)`at `37^(@)` South of West