A car moving along a circular track of radius `50.0m` acceleration from rest at `3.00 ms^(2)` Consider a situation when the car's centripetal acceleration equal its tangential acceleration

Options

A. The angle around the track does the car travel is `1` red
B. The magnitude of the car's total acceleration at that is `3sqrt(2)ms^(-1)`
C. Time elapses before this situation is `sqrt((50)/(3))s`
D. The distance travelled by the car during this time `25m`

Answer

b,c,d

Explanation

Given tangental acceleration
`(dv)/(dt) = 3:v = 3r`
`a_(c) = (v^(2))/(r ) = (9t^(2))/(50)`
`3 = (9.1^(2))/(50) rArr t = sqrt((50)/(3))g`
The angular acceleration of car `alpha = (a)/(r ) = (3)/(50) rad s^(-2)`
The angle rotated by car
`theta = (1)/(2) alpha t^(2) = (1)/(2)(3)/(50)xx(50)/(3)rad^(-1)`
Distance travelled by car upto tjis instant is
`s = theta R = (1)/(2)xx50 = 25m`
Net acceleration of the car is a
`total = sqrt(a_(r)^(2) + a_(t)^(2)) = sqrt(3^(2) + 3^(2)) = 3sqrt(2) ms^(-2)`