physics
posted by george on .
The cheetah can reach a top speed of 114km/h (71mi/h). While chasing its prey in a short sprint, a cheetah starts from rest and runs 50m in a straight line reaching a final speed of 91 km/h.
a.) Determine the cheetah's average acceleration during the short sprint.
b.) Find it's displacement at t=3.0s. (assume the cheetah maintains a constant acceleration throughout the sprint)

change in speed = 91 km/hr = 91,000 m /3600 s
= 25.3 m/s
average speed if a is constant = 12.6 m/s
time for 50 m = 50/12.6 = 3.96 seconds (wow!)
average a = change in speed / time
= 25.3 / 3.96 = 6.4 m/s^2
at 3 seconds
v = a t = 6.4 * 3 = 19.2 m/s
average v for 3 seconds = 19.2/2 = 9.59 m/s
so
distance = 9.59 * 3 = 28.8 meters