(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).

(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

To find the acceleration of the car, a:

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:

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

(b)

We can use again the equation

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,

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

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

one person pulls 90n one way and the other has 100n pulling the other way.

so the net force = 90 - 100 = -10n,

or the ans can be 10n if the picture shows -90n and 100n instead.

distance = 150m

displ = root (50 ^2 + 100^2)

root(2500+10000)

root 125000