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A proud new Jaguar owner drives her car at a speed of 25 m/s into a corner. The coefficients of friction between the road and the tires are 0.70 (static) and 0.40 (kinetic) assuming the car is not skidding while traveling along the curve, what is the magnitude of the centripetal acceleration of the car

Respuesta :

hamzaahmeds

Answer:

ac = 3.92 m/s²

Explanation:

In this case the frictional force must balance the centripetal force for the car not to skid. Therefore,

Frictional Force = Centripetal Force

where,

Frictional Force = μ(Normal Force) = μ(weight) = μmg

Centripetal Force = (m)(ac)

Therefore,

μmg = (m)(ac)

ac = μg

where,

ac = magnitude of centripetal acceleration of car = ?

μ = coefficient of friction of tires (kinetic) = 0.4

g = 9.8 m/s²

Therefore,

ac = (0.4)(9.8 m/s²)

ac = 3.92 m/s²

mickymike92

Based on the data provided, the centripetal acceleration is 3.92 m/s²

What is centripetal acceleration?

Centripetal acceleration is the acceleration of a body moving in a circular path which is directed toward the center of the circle.

In the given question, the frictional force must balance the centripetal force for the car not to skid.

  • Frictional Force = Centripetal Force

where,

Frictional Force = μR

R = mg

F = μmg

Centripetal Force = m

Then

μmg = ma

a = μg

ac = 0.4 * 9.8 m/s²

ac = 3.92 m/s²

Therefore, the centripetal acceleration is 3.92 m/s².

Learn more about centripetal acceleration at: https://brainly.com/question/25243603

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