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12 January, 01:29

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.

(a) What is the magnitude of the car's acceleration, in m/s 2

(b) What average net force magnitude was necessary to stop the car?

(c) Assuming the tires do not skid, what coefficient of static friction between tires and

pavement is needed?

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  1. 12 January, 01:47
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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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