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A 1300-kg car initially has a velocity of 22.2 m/s due south. It brakes to a stop over a 180 m distance.

AWhat is the magnitude of the car’s acceleration, in m/s 2
BWhat average net force magnitude was necessary to stop the car?
CAssuming the tires do not skid, what coefficient of static friction between tires and

pavement is needed?

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(a.) The acceleration of the car can be calculated using the equation,
                                      2ad = v²
Substituting,
                                       2(a)(180 m) = (22.2 m/s)²
From the equation above, we achieve the value of a = 1.369 m/s²

(b) The force necessary is calculated below,
                                   F = (1,300 kg)(1.369 m/s²)
                                      F = 1,779.7 N

(c) Coefficient of friction,
                                      1,779.7 N = (1,300)(9.81)(x)
The value of x from the generated equation,
                                      x = 0.14.

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