Determine the stopping distances for an automobile with an initial speed of 95 km/h and human reaction time of 1.0 s : (a) for an acceleration a = -5.2 m/s2 , (b) for a = -6.6 m/s2 .
Speed of automobile = 95 km/hr
U = 95 km/hr = 95*(5/18) = 26.4 m/sec
V = final speed = 0 m/sec
Now distance traveled by automobile during reaction time will be:
d1 = Velocity*time = (26.4 m/sec)*(1 sec)
d1 = 26.4 m
Part A.
During de-acceleration of a = -5.2 m/sec^2
Using 3rd kinematic equation
V^2 = U^2 + 2*a*S
S = (V^2 - U^2)/(2*a)
S = (0^2 - 26.4^2)/(2*(-5.2)) = 67.0 m
So total distance traveled = d = d1 + S
d = 26.4 + 67.0 = 93.4 m (Ans of part A)
Part B.
During de-acceleration of a = -6.6 m/sec^2
Using 3rd kinematic equation
V^2 = U^2 + 2*a*S
S = (V^2 - U^2)/(2*a)
S = (0^2 - 26.4^2)/(2*(-6.6)) = 52.8 m
So total distance traveled = d = d1 + S
d = 26.4 + 52.8 = 79.2 m (Ans of part B)
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