Couple of physics question

[smileurdentist]

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Sorry, only one question I solved the other one:

A cheetah is hunting. Its prey runs for 2.20 s at a constant velocity of +6.40 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?


Thanks in advance

 

[milski]

Distance for the prey: 2.2 s * 6.4 m/s
Distance for the cheetah: a * 2.2^2 / 2 (d=d0+ t*v0+a*t^2/2, d0 and v0 are 0)

Solving 2.2 * 6.4 = a *2.2^2/2 gives you
6.4 = 1.1 a
or a = 5.9 m/s^2(roughly)