A 1000-kg car is driving toward the north along a straight horizontal road at a speed of 20.0 m/s. The driver applies the brakes and the car comes to a rest uniformly in a distance of 200 m. What are the magnitude and direction of the net force applied to the car to bring it to rest?

Question

Quizplus · Accepted Answer

Using the equation of motion,

Vf^2 = Vi^2 + 2as

where Vf is the final velocity (0 m/s), Vi is the initial velocity (20.0 m/s), a is the acceleration, and s is the distance traveled.

Rearranging the equation to solve for acceleration,

a = -Vf^2 / 2s

Substituting the given values,

a = -(0 m/s)^2 / 2(200 m) = 0 m/s^2

Since the car comes to a rest with uniform acceleration, the net force applied to the car is equal to the frictional force between the wheels and the road, which is given by

Ff = ma

Substituting the given values,

Ff = (1000 kg)(0 m/s^2) = 0 N

Therefore, the magnitude of the net force applied to the car is 0 N.

Since the force acts opposite to the direction of motion (i.e. south), the direction of the net force is south.

The correct answer is therefore option C: 1.00 × 10^10 N south.

A 1000-Kg Car Is Driving Toward the North Along a Straight