(a) -2451 N
We can start by calculating the acceleration of the car. We have:
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](https://tex.z-dn.net/?f=v%5E2%20-%20u%5E2%20%3D%202ad)
To find the acceleration of the car, a:
![a=\frac{v^2-u^2}{2d}=\frac{0-(23.6 m/s)^2}{2(125 m)}=-2.23 m/s^2](https://tex.z-dn.net/?f=a%3D%5Cfrac%7Bv%5E2-u%5E2%7D%7B2d%7D%3D%5Cfrac%7B0-%2823.6%20m%2Fs%29%5E2%7D%7B2%28125%20m%29%7D%3D-2.23%20m%2Fs%5E2)
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](https://tex.z-dn.net/?f=F%3D%281100%20kg%29%28-2.23%20m%2Fs%5E2%29%3D-2451%20N)
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](https://tex.z-dn.net/?f=-1.53%5Ccdot%2010%5E5%20N)
We can use again the equation
![v^2 - u^2 = 2ad](https://tex.z-dn.net/?f=v%5E2%20-%20u%5E2%20%3D%202ad)
To find the acceleration of the car. This time we have
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=\frac{v^2-u^2}{2d}=\frac{0-(23.6 m/s)^2}{2(2 m)}=-139.2 m/s^2](https://tex.z-dn.net/?f=a%3D%5Cfrac%7Bv%5E2-u%5E2%7D%7B2d%7D%3D%5Cfrac%7B0-%2823.6%20m%2Fs%29%5E2%7D%7B2%282%20m%29%7D%3D-139.2%20m%2Fs%5E2)
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](https://tex.z-dn.net/?f=F%3Dma%3D%281100%20kg%29%28-139.2%20m%2Fs%5E2%29%3D-1.53%5Ccdot%2010%5E5%20N)
As we can see, the force is much stronger than the force exerted in part a).