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.
Required:
a. What is the race car's centripetal (radial) acceleration?
b. What is the force responsible for the centripetal acceleration in this case?

Answer 1

Complete 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.

Required:

a. What is the race car's centripetal (radial) acceleration?

b. What is the force responsible for the centripetal acceleration in this case?

O normal

O gravity

O friction

O weight

Answer:

question a

       $a = 16.32 \ m/s^2$

question b

        correct option is option 3

Explanation:

From the question we are told that

   The radius is  $r = 57.0 \ m$

    The constant speed at which the race car is travelling is $v = 30 .5 \ m/s$

Generally from the question we are told that the track is completely flat so the only force pulling the car to the middle is the frictional force hence the centripetal force is due to the frictional force

    Generally the centripetal acceleration is mathematically represented as

      $a = \frac{v^2}{r}$

=>    $a = \frac{30.5^2}{ 57}$

=>    $a = 16.32 \ m/s^2$