Question

A car travels at a constant speed around a circular track whose radius is 2.9 km. the car goes once around the track in 280 s. what is the magnitude of the centripetal acceleration of the car?

Answer 1

centripetal acceleration of the car=a= 1.46 m/s²

Explanation:

The velocity of an object moving around a circular path is given by

$V=\frac{2\pi r}{T}$

T= time=280 s

r= radius of circular track=2.9 km= 2900 m

$V= \frac{2\pi(2900)}{280}$

V=65 m/s

The centripetal acceleration is given by

$a= \frac{v^2}{r}$

$a= \frac{(65)^2}{2900}$

a= 1.46 m/s²