to stop a car first you require a certain reaction time to begin braking; then the car slows under the constant braking deceleration.suppose the total distance moved by the car during these two phases is 56.7 m when its initial speed is 80.5 km/h and 24.4 m when its initial speed is 48.3 km/h. What are:

initial speed in first case, u1 = 80.5 kmph  = 22.36 m/s

total distance moved to stop in this case, s1 = 56.7 m

 

initial speed in second case, u2 = 48.3 kmph = 3.41 m/s

total distance moved to stop in this case, s2 = 24.4 m

 

finail velocity, v  = 0 m/s

 

let the reaction time be T

and magnitude of deceleration be a

 

thus if the time taken to stop the car in first case is t1 and in second case is t2

thus using first equation of motion,

t1 = - u1/a            ...............(1)

t1 = - u2/a            ...............(2)

and using third equation of motion,

u1²/2.a + u1.T = s1                 ............(3)

 

similarly for the second case:

 u2²/2.a + u2.T = s2                 ............(4)

 

thus now solving eq. 3 and 4 simultaneously, we can find T and a