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 m57.0 m . If the track is completely flat and the race car is traveling at a constant 31.5 m/s31.5 m/s (about 7.0×101 mph7.0×101 mph ) around the turn, what is the race car's centripetal (radial) acceleration?

Answer 1

Answer:

Centripetal acceleration of the car is 17.4 m/s²

Explanation:

It is given that,

Radius of the track, r = 57 m

Speed of car, v = 31.5 m/s

We need to find the centripetal acceleration of the race car. The formula for the centripetal acceleration is given by :

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

$a=\dfrac{(31.5\ m/s)^2}{57\ m}$

$a=17.4\ m/s^2$

So, the centripetal acceleration of the race car is 17.4 m/s². Hence, this is the required solution.