Cheetahs can accelerate to a speed of 20.0 m/s in 2.50 s and can continue to accelerate to reach a top speed of 30.0 m/s. Assume the acceleration is constant until the top speed is reached and is zero thereafter. Let the +x-direction point in the direction the cheetah runs. Express the cheetah’s top speed (vtop) in miles per hour (mi/h). Starting from a crouched position, how much time (t-accel) does it take a cheetah to reach its top speed, and what distance (d) does it travel in that time?

The cheetah’s initial velocity is 0 m/s. Let’s use the following equation to determine the acceleration.

Explanation:

Acceleration = (v/t) = (20/2.5) = 8m/sec^2.

1) Time to 29m/sec. = (v/a) = (29/8) = 3.625 secs. (3.6)

Distance covered to top speed = (v^2/2a) = 52.5625 metres (52.6).

2) Distance to run at top speed = (120 - 52.5625) = 67.4375 metres.

Time = (d/v) = (67.4375/29) = 2.325 secs.

Total time to warthog = (2.325 + 3.625) = 5.95 secs.

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