Posted by **george** on Sunday, February 3, 2013 at 6:11pm.

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)

- physics -
**Damon**, Sunday, February 3, 2013 at 7:47pm
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

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