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

Answer:

$a=16.32\ m/s^2$

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 :

$a=\dfrac{v^2}{r}\\\\a=\dfrac{(30.5)^2}{57}\\\\=16.32\ m/s^2$

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