Question

(a) Calculate the force needed to bring a 950-kg car to rest from a speed of 90.0 km/h in a distance of 120 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

Answer 1

(a)

mass of the car: m=950 kg

Initial speed: $u = 90.0 km/h=25 m/s$

Final speed: $v=0$ (the car comes to rest)

distance: $S=120 m$

We can find the acceleration by using the following SUVAT equation:

$v^2 -u^2 =2aS$

Re-arranging it and replacing the numbers, we find the acceleration

$a=\frac{v^2-u^2}{2S}=\frac{0-(25 m/s)^2}{2(120 m)}=-2.6 m/s^2$

So now we can calculate the force using Newton's second law:

$F=ma=(950 kg)(-2.6 m/s^2)=-2470 N$

And the negative sign means the force is applied against the direction of motion.

(b)

In this case, the distance is different:

$S=2.00 m$

so, the acceleration in this case is

$a=\frac{v^2-u^2}{2S}=\frac{0-(25 m/s)^2}{2(2 m)}=-156.3 m/s^2$

And so, the force applied in this case is

$F=ma=(950 kg)(-156.3 m/s^2)=-148 500 N$

which is much larger than the force exerted in the previous exercise.