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
