A jeep runs around a curve of radius `0.3` km at a constant speed of `60ms^(-1)`. The jeep covers a curve of `60^(@)` arc

Options

A. resultant change in velocity of jeep is `60ms^(-1)`
B. instantaneous acceleration of jeep is `12ms^(-2)`
C. average acceleration of of jeep is approximately `11.5ms^(-2)`
D. All are correct

Answer

D

Explanation

`Deltat=("Distance travelled")/("Speed")`
`=((2piR//6))/(v)=(3.14xx300)/(60xx3)=5.23s`

(i) `|Deltav|=|v_(f)|-|v_(i)|`
`=sqrt(v^(2)+v^(2)=2"v.v cos 60"^(@))="2 v sin 30"^(@)="60 ms"^(-1)`
(ii) `a_(i)=(v^(2))/(R)=12ms^(-2)`
(iii) `|a_(av)|=(|Delta|)/(Deltat)=(60)/(5.23)=11.5ms^(-2)`