Question

A car of mass 1100 kg is traveling at 24 m s−1. The driver applies the brakes and the car decelerates uniformly and comes to rest in 20 s.

a) Calculate the change in momentum of the car.
b) Calculate the braking force on the car.
c) Determine the braking distance of the car.

Answer 1

Answer:

A car of mass 1100 kg is traveling at 24 m s−1. The driver applies the brakes and the car decelerates uni

Answer 2

The change in momentum is -13200 kg m/s. The braking force is  -660 N. The braking distance of the car is 961.67 m.

To Find:

a) Change in momentum of the car

b) Braking force on the car

c) Braking distance of the car.

Given:

• Mass of the car: 1100 kg
• Initial velocity of the car: 24 m/s
• Time taken to come to rest: 20 s

Solution:

a) Change in momentum is given by the equation Δp = m * Δv, where Δv is the change in velocity and m is the mass of the object.

Since the car decelerates uniformly from 24 m/s to 0 m/s over 20 s, the average change in velocity is (-24 m/s - 0 m/s)/2 = -12 m/s.

Therefore, the change in momentum is Δp = 1100 kg * -12 m/s = -13200 kg m/s.

b) The braking force can be calculated using Newton's second law of motion, F = m * a, where F is the force, m is the mass of the object and a is the acceleration.

The acceleration of the car can be calculated as a = Δv / t, where t is the time it takes to stop.

So, a = -12 m/s / 20 s = -0.6 m/s^2.

Therefore, the braking force is F = 1100 kg * -0.6 m/s^2 = -660 N.

c) Braking distance can be calculated using the formula d = v^2 / (2 * a), where v is the initial velocity and a is the acceleration.

Substituting v = 24 m/s and a = -0.6 m/s^2, we get

d = 24^2 / (2 * -0.6) = 1152 m/s^2 / -1.2 m/s^2 = 961.67 m.

So, the braking distance of the car is 961.67 m.

Therefore,The change in momentum is -13200 kg m/s. The braking force is  -660 N. The braking distance of the car is 961.67 m.

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