A car travelling at a speed of `30 ms^(-1)` due north along the highway makes a left turn on to a sied road which heads towards due west. It takes `40 s` for the car to complere the turn. At the end of `40 s`, the caar has a speed of `20 ms^(-1)` along the side road. Derermime the magnitude of average acceleration over the `40 second` interval.

Explanation

Refer to figure, initial velocity, `vecv_(1)=overset(-)(OA)=30ms^(-1)`, due north. Final velocity `vecv_(2)=overset(-)(OB)=20ms^(-1)` due west.
Change in velocity `vecv_(2)-vecv_(1)=overset(-)(OB)-overset(-)(OA)=overset(-)(AB)`
`|vec_(2)-vecv_(1)|=AB sqrt((OA)^(2)+(OB)^(2))`
`=sqrt(30^(2)+20^(2))=sqrt(1300)=36ms^(-1)`
Average acceleration `a_(av)=(|vecv_(2)-vecv_(1)|)/(t)=36/40=0.9 ms^(-2)`