Answered

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 :

Answer:

[tex]t = 3.49 s[/tex]

[tex]a = -5.02 m/s^2[/tex]

Explanation:

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

[tex]v_i = 17.5 m/s[/tex]

finally it comes to rest so final speed is given as

[tex]v_f = 0[/tex]

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

[tex]d = 30.5 m[/tex]

now by equation of kinematics we know that

[tex]d = (\frac{v_f + v_i}{2})t[/tex]

here we have

[tex]30.5 m = (\frac{0 + 17.5}{2}) t[/tex]

[tex]30.5 = 8.75 t[/tex]

[tex]t = 3.49 s[/tex]

Now in order to find the acceleration we know that

[tex]v_f - v_i = at[/tex]

[tex]0 - 17.5 = a(3.49)[/tex]

[tex]a = -5.02 m/s^2[/tex]

Other Questions