Question

A driver traveling at 100 km/h on a rural road sees a sheep 32 m ahead in the middle of the road. His reaction time is 0.70 s, after which he breaks at an acceleration of -5.88 m/s2. Unfortunately, this is not sufficient to avoid hitting the sheep. At what speed does the driver end up hitting the sheep?

Answer 1

Step-by-step explanation:

100 km/hr = 27.78 m/s

in .7 sec the car will travel 19.44 meters

leaving only 32 m - 19.44m = 12.56 m to deccelerate

vf = vo + at

= 27.78 m/s + (-5.88 m/s^2) t <==== we must find 't'

df = do + vo t + 1/2 a t^2

12.56 = 0 + 27.78 t - 1/2 (5.88) t^2 re-arrange;

0 = - 2.94 t^2 + 27.78 t - 12.56 Use Quadratic Formula to find t = .476 s

Now

vf = vo + at

= 27.78 - 5.88(.476) = 24.98 m/s when the sheep get muttoned

= 89.9 km/hr