A car going due North at `10sqrt(2) ms^(-1)` turns right through an angle of `90^(@)` without changing speed. The change in velocity of car is

Options

A. `20 ms^(-1)` in South -East direction
B. `20sqrt(20) ms^(-1)` in South -East direction
C. `20 ms^(-1)` in North-West dircrection
D. `20 ms^(-1)` in North-West direction

Answer

A

Explanation

(a) initial velocity is `v_(i)=10sqrt(2) hat(j)`

Final velocity is `v_(f)= 10sqrt(2) hat(j)`
`:.` The change in velocity is

`DeltaV= v_(f)-v_(i)= 10sqrt(2)hat(i)-10sqrt(2)hat(j)`
`:. |Delta v|= sqrt((10sqrt(2))^(2)+(10sqrt(2))^(2))=sqrt(400)`
`=20 ms^(-1)` in South- East direction.