Question

The "reaction time" of the average automobile driver is about 0.7 s . (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 11.6 ft/s2.

Compute the total distance covered in coming to a stop after a signal is observed from an initial velocity of 14.4 mi/h . (in a school zone)

Compute the total distance covered in coming to a stop after a signal is observed from an initial velocity of 51.4 mi/h.

Answer 1

Right. Let me do some conversions first.
14.4 m/h = 21.12 ft/ s
51.4 m/h = 75.38 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 = 21.12 * 0.7 = 14.784 ft

v = 0; u = 21.12; a = - 11.6 so

0 = 21.12^2 - (2 * 11.6 * s)
s = 21.12^2 / 23.2 = 19.22 ft

So total stopping distance from 15 m/h = 19.22 + 14.784 = 34.004 ft

b) The distance travelled during the reaction time = 75.38 * 0.7 = 52.766 ft

v = 0; u = 75.38; a = - 11.6
0 = (75.38)^2 - (2 * 11.6 * s)
s = (75.38^2) / 23.2 = 244.92 ft

So total stopping distance from 51.4 m/s = 244.92 + 52.766 = 297.686ft