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Chapter 02, Problem 79 Consult Concept Simulation 2.1 for help in preparing for this problem. A cheetah is hunting. Its prey runs for 2.50 s at a constant velocity of +9.70 m/s. Starting from rest, what constant acceleration must the cheetah maintain in order to run the same distance as its prey runs in the same time?

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Answer #1

Time t = 2.5 s

Constant velocity v = 9.7 m/s

Distance travel in time t is S = vt = 9.7 x 2.5 = 24.25 m

For cheetah :

Initial velocity u = 0

Time t= 2.5 s

Distance S = 24.25 m

From the relation S = ut +(1/2)at 2

24.25 = 0 +[0.5xax2.5 2]

= 3.125 a

a = 7.76m/s 2

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