A body moving in a curved path possesses a velicty `3m//s ` towards north at any instant of its motion . A fter 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 arfe drawn from a common origin . The vecto difference of 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`
The direction is given by
`tan theta =3/4=0.75, theta =37^(0)`
Average acceleration `=|vecv-vecv_(0)|/t=5/(10)=0.5m//s^(2)`at`37^(@)`south of west