A 825 -kg car comes to rest from a speed of 77.5 km/h in a distance of 130 m. As

Question

A 825-kg car comes to rest from a speed of 77.5 km/h in a distance of 130 m. Assume the car is initially moving in the positive direction.

1) 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.

2) 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.

3) What is the ratio of the force on the car from the concrete to the braking force?

Explanation / Answer

Given:-

m= 825 kg

vf=0 m/s

vi= 77.5 km/h= 21.53 m/s [ 1km/h= 5/18 m/s]

d=130m

1) Car come to a stop

Using the kinematic equation, vf2= vi2+2ad

we get a=-1.78 m/s2

F net = ma [apply Newton second law]

we get , F net= -1470.55N { direction of force is opposite due to decceleration}----------F braking force

2) Car hits a concrete abutment

vi= 77.5 km/h= 21.53 m/s [ 1km/h= 5/18 m/s]

vf=0 m/s

d =2m

Using the kinematic equation, vf2= vi2+2ad

we get a=-115.86 m/s2 64.99~ 65

F net = ma [apply Newton second law]

we get , F net= -9.56*104N { direction of force is opposite due to decceleration}----------------F concrete

3) Ratio= F concrete/F braking force= -9.56*104N/ -1470.55N= 65

F concrete = 65* F braking force

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