Question

Suppose a NASCAR race car rounds one end of the Martinsville Speedway. This end of the track is a turn with a radius of approximately 57.0 m . If the track is completely flat and the race car is traveling at a constant 30.5 m/s (about 68 mph ) around the turn, what is the race car's centripetal (radial) acceleration

Answer 1

`a=16.32\ m/s^2`

Step-by-step explanation:

Given that,

The radius of the track, r = 57 m

The speed of a race car, v = 30.5 m/s

We need to find the centripetal acceleration of the car. Its formula that is use to find it is given by :

`a=(v^2)/(r)\\\\a=((30.5)^2)/(57)\\\\=16.32\ m/s^2`

So, the car's centripetal acceleration is
`16.32\ m/s^2`.