The \"reaction time\" of the average automobile driver is about 0.7s . (The reac

Question

The "reaction time" of the average automobile driver is about 0.7s . (The reaction time is the interval between the perception of a signal to stop and the application of the brakes.) An automobile can slow down with an acceleration of 13.6ft/s2

A) Compute the total distance covered in coming to a stop after a signal is observed from an initial velocity of 12.6mi/h . (in a school zone) (Answer in ft)

B)Compute the total distance covered in coming to a stop after a signal is observed from an initial velocity of 55.6mi/h . (Answer in ft)

Explanation / Answer

Let me do some conversions first.
12.6 m/h = 12.6 * 5280 / 3600 = 18.48 ft/ s
55.6 m/h = 81.55 ft / s

I'll be using the equation of motion that connects initial velocity, final velocity, acceleration and distance.

v^2 = u^2 - 2 a s
where v = final velocity (ft/s); u = initial velocity (ft / s); a = acceleration (ft / s^2); s = distance (ft)
Since it is slowing down, the acceleration is negative.

a) The distance travelled during the reaction time = 18.48 * 0.7 = 12.94 ft

v = 0; u = 22; a = - 13 so

0 =18.48^2 - (2 * 13 * s)
s = 18.48^2 / 26 =13.135 ft

So total stopping distance from 12.6 m/h = 13.135 + 12.94 = 26.075 ft

b) The distance travelled during the reaction time = 81.55 * 0.7 = 57.08 ft

v = 0; u = 81.55; a = - 13
0 = (81.55)^2 - (2 * 13 * s)
s = (81.55^2) / 26 = 255.78 ft

So total stopping distance from 55.6 m/s = 255.78 + 57.08 = 312.16 ft

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