physics
posted by tommy on .
A driver in a car traveling at a speed of 70 mi/h sees a deer 90 m away on the road. Calculate the minimum constant acceleration that is necessary for the car to stop without hitting the deer (assuming that the deer does not move in the meantime).

Vo = 70 mph = 102.7 ft/s = 31.3 m/s
The average speed while decelerating must be Vo/2 = 15.65 m/s
Deceleration time = 90 m/(avg. speed) = 5.75 s
Deceleration rate = Vo/(deceleration time) = 31.3/5.75 = 5.44 m/s^2
Check: distance travelled while decelerating =
Vo*t  (a/2)t^2 = 31.3*5.75  (2.72)(5.75)^2 = 90 m