Question

A 950-kg car comes to rest from a speed of 82.5 km/h in a distance of 120 m. Assume the car is initially moving in the positive direction.Part a) If the brakes are applied consistently and 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

Using the kinematics equations:

a)

`\begin{gathered} v^2=v_o^2+2a(d) \\ where: \\ v_o=82.5km/h \\ d=120m \\ v_=0 \\ so: \\ a=-(v_o^2)/(2d)=(22.92^2)/(2(120)) \\ a=-2.19m/s^2 \end{gathered}`

Now, using newton's second law:

`\begin{gathered} F=ma \\ F=950(-2.19) \\ F=-2079.417N \end{gathered}`

(b)

Using the same equation we use previously:

`a=-(v_o^2)/(2d)=-(22.92^2)/(2(2))=-131.3316m/s^2`

Using newton's second law again:

`\begin{gathered} F=ma \\ F=950(-131.3316) \\ F=-124765.02N \end{gathered}`

(c)

`-(124765.02)/(-2079.417)=60`

The ratio is 60.