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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?

Respuesta :

Answer:

[tex]a_c = 13.26 m/s^2[/tex]

Friction Force

Explanation:

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

so we know that

[tex]a_c = \frac{v^2}{R}[/tex]

so here we have

[tex]v = 27.5 m/s[/tex]

[tex]R = 57 m[/tex]

so we have

[tex]a_c = \frac{27.5^2}{57}[/tex]

[tex]a_c = 13.26 m/s^2[/tex]

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.

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