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:

Asked by susmita bhattacharjee | 2nd Jun, 2013, 06:49: PM

Expert Answer:

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

Answered by  | 2nd Jun, 2013, 11:11: PM

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