The "reaction time" of the average automobile driver is about 0.700 s. (The reaction time is the interval between the perception of a signal to stop and the application of the brakes.) If an automobile can slow down with an acceleration of 12.0 ft/s^{2}, compute the total distance covered in coming to a stop after a signal is observed (a) from an initial velocity of 15.0 mi/h (in a school zone) and (b) from an initial velocity of 55.0 mi/h.

Question

The "reaction time" of the average automobile driver is about 0.700 s. (The reaction time is the interval between the perception of a signal to stop and the application of the brakes.) If an automobile can slow down with an acceleration of 12.0 ft/s^{2}, compute the total distance covered in coming to a stop after a signal is observed (a) from an initial velocity of 15.0 mi/h (in a school zone) and (b) from an initial velocity of 55.0  mi/h.

drwls · Accepted Answer

The total stopping distance is:

X = Vo*(reaction time) + (a/2)*(deceleration time)^2

Reaction time = tr = 0.70 s
Deceleration time = Vo/a

X = Vo*tr + (1/2)(Vo^2)/a

(a) If Vo = 15 mi/h = 22 ft/s,
X = 22*(0.700) + (1/2)(22/15)^2/0.12
= 15.4 + 6.1 = 21.5 ft

(b) repeat calculation for Vo = 55 mi/h = 80.67 ft/s