Question

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 .

Answer 1

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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