# Find Acceleration with only Initial Velocity and Final position?

1. Jun 10, 2009

### fireykitty

1. The problem statement, all variables and given/known data

A hockey puck sliding on a frozen lake comes to rest after 238 m. If its initial velocity is 3.8 m/s what is its acceleration if it is assumed constant? Answer in units of m/s^2

How long is it in motion? What is its speed after traveling 180 m?

2. Relevant equations

x=xo + vot +1/2at^2

3. The attempt at a solution

v^2=(vo)^2+2a(x-xo)
...v^2=(3.8)^2 + 2a (-238)

but i don't know the velocity, cause i don't have the time!

3.8 m/s * 1/238 m = time?

I find that I can't do anything without the time! I'm so lost and confused. PLEASE HELP.

Last edited: Jun 10, 2009
2. Jun 10, 2009

### diazona

Here's a hint: what does "comes to rest" mean? ;-)

3. Jun 10, 2009

### LowlyPion

Welcome to PF.

That equation works. Your final velocity is 0 so ...

v2 = 3.82 = 2*a*238

Then you have your acceleration a.

With that v = a*t so t becomes easy to find.

To find the speed at 180 m just use the equation you gave above to determine the speed at that distance.

4. Jun 11, 2009

### fireykitty

But then we have two variables, a, and v, how do I solve then? it doesn't work if you manipulate and substitute v =at because then the other variable, t, comes in.

5. Jun 11, 2009

### diazona

No you don't, you only have one, a. Look carefully at the equation LowlyPion wrote for you.

6. Jun 11, 2009

### fireykitty

ok, i tried it, solved for a and I got .030 m/s^2 = a.

i typed that into my homework answers, and it said it was incorrect (we have multiple times to try it)

and when I do v=at,

3.8=(.030)t,

I get 125.3 seconds....seems a bit too much time. am i doing something wrong?

7. Jun 11, 2009

### LowlyPion

I trust you entered it as a = -.03 m/s2

It is decelerating after all.