Question

(a) Calculate the force (in N) needed to bring a 1100 kg car to rest from a speed of 85.0 km/h in a distance of 125 m (a fairly typical distance for a non-panic stop). (b) Suppose instead the car hits a concrete abutment at full speed and is brought to a stop in 2.00 m. Calculate the force exerted on the car and compare it with the force found in part (a).

Answer 1

(a) -2451 N

We can start by calculating the acceleration of the car. We have:

`u=85.0 km/h = 23.6 m/s` is the initial velocity

v = 0 is the final velocity of the car

d = 125 m is the stopping distance

So we can use the following equation

`v^2 - u^2 = 2ad`

To find the acceleration of the car, a:

`a=(v^2-u^2)/(2d)=(0-(23.6 m/s)^2)/(2(125 m))=-2.23 m/s^2`

Now we can use Newton's second Law:

F = ma

where m = 1100 kg to find the force exerted on the car in order to stop it; we find:

`F=(1100 kg)(-2.23 m/s^2)=-2451 N`

and the negative sign means the force is in the opposite direction to the motion of the car.

(b)
`-1.53\cdot 10^5 N`

We can use again the equation

`v^2 - u^2 = 2ad`

To find the acceleration of the car. This time we have

`u=85.0 km/h = 23.6 m/s` is the initial velocity

v = 0 is the final velocity of the car

d = 2.0 m is the stopping distance

Substituting and solving for a,

`a=(v^2-u^2)/(2d)=(0-(23.6 m/s)^2)/(2(2 m))=-139.2 m/s^2`

So now we can find the force exerted on the car by using again Newton's second law:

`F=ma=(1100 kg)(-139.2 m/s^2)=-1.53\cdot 10^5 N`

As we can see, the force is much stronger than the force exerted in part a).