Answer:
[tex] 1.51 * 10^4 N [/tex]
Explanation:
The following parameters are given
initial speed u = 22.8 m/s
final speed v = 2.9 m/s
distance s = 5.55 m
mass of car m = 328 kg
deceleration a= ??
Using Newton's third equation of motion
[tex] v^2 = u^2 -2as [/tex]
Here, we chose -a since the car is decelerating (negative acceleration)
Hence [tex] 2.9^2 = 22.8^2 - (2*a*5.550) [/tex]
Solving the equation, we obtain a = 46.074 [tex] m/s^2 [/tex]
Hence the braking force = mass of car * deceleration
= 328 * 46.074
= [tex] 1.51 * 10^4 N [/tex]
