A 775-kg car comes to rest from a speed of 95 km/h in a distance of 130 m. Assume the car is initially moving in the positive direction.

Question

Part (a) If the brakes are the only thing making the car come to a stop, calculate the force (in newtons, in a component along the direction of motion of the car) that the brakes apply on the car.

Part (b) Suppose instead of braking that the car hits a concrete abutment at full speed and is brought to a stop in 2.00 m. Calculate the force, in newtons, exerted on the car in this case.

Part (c) What is the ratio of the force on the car from the concrete to the braking force?

Answer 1

a). In order for the car to stop it must decelerate (negatively accelerate). The acceleration needed to bring the car from an initial velocity of 95 km/h (26.3 m/s) to a final velocity of 0 m/s in 130 m can be found using the equation:

V_f² = V_o² + 2ad, Solving for a gives us a =(V_f² - V_o² ) / 2d = -2.66 m/s² (negative because it is decelerating)

using Newton's 2nd Law to find the force required to cause that deceleration

F=ma = 775 x -2.66 = -2060 N the negative force meaning it is in the opposite direction as the motion.

b). Substituting the new stopping distance in to what you did above (2.00 m instead of 130 m) will give you an acceleration of -173 m/s² and a force of -134,000 N.

c). The ratio of the concrete force to the braking force would be (in 2 significant figures):

-134,000 N / -2060 N = 65