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 27.5 m/s (about 62 mph ) around the turn, what is the race car's centripetal (radial) acceleration?

What is the force responsible for the centripetal acceleration in this case?
friction, normal, gravity, or weight?

Answer 1

`a_c = 13.26 m/s^2`

Friction Force

Step-by-step explanation:

As we know that centripetal force is the product of mass and centripetal acceleration

so we know that

`a_c = (v^2)/(R)`

so here we have

`v = 27.5 m/s`

`R = 57 m`

so we have

`a_c = (27.5^2)/(57)`

`a_c = 13.26 m/s^2`

This acceleration is given by the force which may be towards the center of the circular path

Here in the above case it is possible due to friction force.