7. A car is moving at 50.0 mph when the driver applies brakes. Determine the distance it covers before coming to a halt. Coefficient of static friction between the tires and surface of
the road is 0.514. Mass of the car is 1000 kg.

The car will cover 49.65 m distance before stopping due to application of brake.

Given data:

The mass of car is, M = 1000 kg.

The initial speed of car is, u = 50.0 mph = (50)(0.447 m/s) = 22.35 m/s.

The coefficient of static friction is, .

The given problem can be solved using the third kinematic equation of motion. Which is,

....................................................(1)

Here, v is the final speed, v = 0 (Because car is finally stopping)

s is the distance covered before stopping and a is the magnitude of acceleration.

Now, since frictional force opposes the motion. Then,

Substituting the values in equation (1) as,

Thus, the car will cover 49.65 m distance before stopping due to application of brake.

Learn more about the Kinematic equations of motion here:

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The distance to come to a halt is approximately 49.53 meters

Explanation:

Thee given parameters are;

The speed of the car, v = 50 mph = 22.35 m/s

The mass of the car, m = 1000 kg

The coefficient of friction,  = 0.514

The force of friction of the brake = Mass × Gravity × Friction = 1000 × 9.81 × 0.514 = 5042.34 N

The initial kinetic energy of the car = 1/2×m×v² = 1/2 × 1000 × 22.35² = 249761.25 J

The work done by the brake = Force of the brake × Distance, d, to come to halt

By conservation of energy, we have;

The work done by the brake = The initial kinetic energy of the car

∴ The initial kinetic energy of the car = Force of the brake × Distance, d, to come to halt

The initial kinetic energy of the car = 249761.25 J = 5042.34 N × Distance, d, to come to halt

∴The distance to come to a halt = 249761.25 J /(5042.34 N) ≈ 49.53 meters

The distance to come to a halt ≈ 49.53 meters.

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