Question

A 1400 kg car driving at 25 m/s slams on its brakes. The coefficient of kinetic friction between the tires and the road is 0.7.

Answer 1

The acceleration of the car is 6.86 m/s² and the time taken for the car to stop is 3.64 s.

The given parameters;

• mass of the car, m = 1400 kg
• Initial velocity of the car, u = 25 m/s
• coefficient of kinetic friction, μ = 0.7

The acceleration of the car is calculated as follows;

a = μg

a = 0.7 x 9.8

a = 6.86 m/s²

The time taken for the car to stop is calculated by using Newton's second law of motion;

F = ma

$F = \frac{mv}{t} \\\\ma = \frac{mv}{t}\\\\a = \frac{v}{t} \\\\t = \frac{v}{a} \\\\t = \frac{25}{6.86} \\\\t = 3.64 \ s$

Thus, the acceleration of the car is 6.86 m/s² and the time taken for the car to stop is 3.64 s.

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