A car is travelling towards east at `10m*s^(-1)`. It takes 10s to change its direction of motion to north and continues with the same magnitude of velocity. Find the magnitude and direction of the average acceleration of the car.

Explanation

In Fing2.24., change in velocity of the car =final velocity -initial velocity
=`(10 m*s^(-1)` towards onrth) -(`10 m *s^(-1)` towards east)
`=vec(AB)-vec(OA)=vec(AB)+vec(AO)=vec(AC)`
[from parallelogram law of vector addition ]
`therefore AC^(2)=AB^(2)+BC^(2)=AB^(2)+AO^(2)`
`=10^2+10^2=100+100=200`
`therefore AC=10sqrt(2)m*s^(-1)`

`therefore` Average acceleration = `("change in velocity")/("time")`
`=(10sqrt(2))/(10)=sqrt(2) m*s^(-2)`
it is direction along `vec(AC)` that makes angle `theta` with `vec(AB)`, and `tan theta =(BC)/(AB)=(AO)/(AB)=(10)/(10)=1=tan 45^(@)`
`therefore theta =45^(@)`
Hence , average acceleration is direction towards north-west.