What average braking force is required to stop a 1134-kg car traveling at a speed of 83 km/hr before it reaches a stop sign that is 98 m away?

Given:

m(mass of the car)=1134 Kg

u(Initial velocity)=83Km/HR=23m/s

s(distance traveled by the car)=98m

v(final velocity)=0(as it is given the car stops).

Now we know,

v=u+at

Where v is the final velocity

u is the initial velocity

a is the acceleration

t is the time

0=23+at

at=-23

Also

s=ut+1/2(at^2)

s is the distance covered by the car

u is the initial velocity

t is the time necessary for the car to cover a particular distance.

a is the acceleration

Now substituting these values we get

98=23t-1/2(23t)

98=23t-11.5t

11.5t=98

t=8.52secs

Now we have already derived

at=-23

ax8.52=-23

a=-23/8.52

a=-2.75 m/s^2

F=mxa

Where F is the force acting on the car.

m is the mass of the car.

a is the acceleration.

F=1134 x-2.75

F=-3119N

it about 600 mph from what i calculated

a is the best answer your welcome