What will be the average braking force needed to bring a car of mass 800kg traveling 72km/hr to a stop at a distance of 100m

F = m*a

First, we find the acceleration.

Use the following equation of motion:
2*a*s = v^2 - u^2
s = 100m, v = 0, u = 72km/hr
= 20m/s

=> 2*a*100 = 0 - (20)^2
=> a = -400/200
= -2m/s^2

The sign is negative because the acceleration is retarding in nature. Similarly, the force will also have a negative sign:

F = m*a
= 800*(-2)
= -1600N

Hence, the magnitude of the force required will be 1600N

and, there is another method.

starting velocity=20m/s, ending velocity=0, average velocty=10m/s
time to go 100m, time=distance/avgV=10sec
Vf=vi+at
0=20)a*10
a=-2m/s then us F=ma to get force.

To find the average braking force needed to bring a car to a stop, we can use the formula:

F = (m * v^2) / (2 * d),

where F is the braking force, m is the mass of the car, v is the initial velocity, and d is the distance traveled.

First, let's convert the velocity from km/hr to m/s. We can do this by dividing by 3.6:

v = 72 km/hr / 3.6 = 20 m/s.

Now, substituting the given values into the formula:

F = (800 kg * (20 m/s)^2) / (2 * 100 m).

Simplifying further:

F = (800 kg * 400 m^2/s^2) / 200 m = 160,000 N / 200 m.

F = 800 N.

Therefore, the average braking force needed to bring the car to a stop is 800 Newtons.