# Answer to Question #1024 in Mechanics | Relativity for sabeen

Question #1024

A cheetah is traveling across the African countryside, initially moving at a constant velocity of 8.0 m/s. It sees a gazelle at t = 0 seconds and begins to accelerate in a straight line. At t = 4.0 seconds it has a velocity of 23 m/s and then continues to move at this constant velocity. At t = 5.0 seconds, the cheetah is standing still, having knocked over the gazelle.(a) What is the average acceleration of the cheetah from t = 0 to t = 5.0 seconds?

Expert's answer

v = v

a = 3.75 m/s^2

From t=4s to 5s the acceleration is equal to zero, thus the average is:

int(a(t)dt) / T = 4*3.75 / 5 = 3 m/s

_{0}+ at = 8 + 4*a = 23a = 3.75 m/s^2

From t=4s to 5s the acceleration is equal to zero, thus the average is:

int(a(t)dt) / T = 4*3.75 / 5 = 3 m/s

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