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 24.5 m/s24.5 m/s (about 55 mph55 mph ) around the turn, what is the race car's centripetal (radial) acceleration

Answer 1

Answer:

10.53m/s²

Explanation:

Centripetal acceleration is the acceleration of an object about a circle. The formula for calculating centripetal acceleration is expressed by:

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

v is the velocity of the car = 24.5m/s

r is the radius of the track = 57.0m

Substitute the given values into the formula:

$a = \frac{24.5^2}{57} \\\\a = \frac{600.25}{57}\\ \\a = 10.53m/s^{2}$

Hence the centripetal acceleration of the race car is 10.53m/s²