Constant Acceleration of a cheetah

[12345ME]

Homework Statement

A cheetah is hunting. Its prey runs for 3.2 s at a constant velocity of +10.0 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?

Homework Equations

V=Vo + a(t)

The Attempt at a Solution

I plug in the numbers to get 3.1 m/s but i know I am missing something. Help with answer would be appreciated!

 

[rl.bhat]

What is the distance traveled by the cheetah?
Use the kinematic equation d = vo*t + 1/2*a*t^2 to find a.

 

[12345ME]

this is all the problem says, doesn't give a distance.

 

[rl.bhat]

this is all the problem says, doesn't give a distance.

Velocity of the cheetah is given, time taken be it is given. What is the relation between distance, velocity and time?

 

[rootX]

A cheetah is hunting. Its prey runs for 3.2 s at a constant velocity of +10.0 m/s.

What distance does the prey runs?

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?

You know the d now, t is 3.2s ... find a

 

[12345ME]

still have no idea what the distance is

 

[rl.bhat]

still have no idea what the distance is

Open any textbook and find the relation between velocity, distance and time.

 

[12345ME]

distance= velocity x time right?

 

[rl.bhat]

distance= velocity x time right?

Right.

 

[12345ME]

but when you use d = vo*t + 1/2*a*t^2

i get 0= 1/2*a*3.2^2

which equals 0

 

[12345ME]

lost

 

[rl.bhat]

lost

d = 3.2 s*10 m/s

 

[12345ME]

d = 3.2 s*10 m/s

ya but so does Vo*t on the right side of the equation so they cancel each other out

 

[12345ME]

it comes out to
32=32 + 1/2*a*3.2^2
0=1/2*a*3.2^2
?

 

[rl.bhat]

it comes out to
32=32 + 1/2*a*3.2^2
0=1/2*a*3.2^2
?

Initial velocity of cheetah is zero.

 

[12345ME]

6.25 thank you!