Question

A cheetah running at 17.5 m/s looses its prey and begins to slow down at a constant acceleraton, fianlly stopping 30.5 m later. How long does it take the cheetah to come to rest and what is the cheetah's acceleration while doing so? I know the answer which is ∆t= 3.49 s but want to see the work how it gets 3.49 s.

Answer 1

`t = 3.49 s`

`a = -5.02 m/s^2`

Step-by-step explanation:

As we know that initial speed of the cheetah is given as

`v_i = 17.5 m/s`

finally it comes to rest so final speed is given as

`v_f = 0`

now we know that distance covered by cheetah while it stop is given as

`d = 30.5 m`

now by equation of kinematics we know that

`d = ((v_f + v_i)/(2))t`

here we have

`30.5 m = ((0 + 17.5)/(2)) t`

`30.5 = 8.75 t`

`t = 3.49 s`

Now in order to find the acceleration we know that

`v_f - v_i = at`

`0 - 17.5 = a(3.49)`

`a = -5.02 m/s^2`