# Constant acceleration of skateboarder

1. Sep 15, 2008

### southernbelle

1. The problem statement, all variables and given/known data
A skateboarder coasts down a long hill for 6.00 seconds with constant acceleration, and starting from rest, covers a distance x1. On a second trip down the same hlil starting fron rest, he coasts for 2.00 seconds, covering a distance x2. Find the ratio of x1 to x2.

2. Relevant equations

3. The attempt at a solution
To me the obvious answer would be
6s to 2s, so 3 to 1.
But something tells me it's not that simple.
What am I missing?

2. Sep 16, 2008

### Redbelly98

Staff Emeritus
Welcome to Physics Forums southernbelle.

Correct, it's not that simple. That 3:1 ratio would work for a constant velocity (no acceleration).

What equations do you know about that deal with constant acceleration?

3. Sep 16, 2008

### southernbelle

I know that xf=xi +vi(t) + 1/2at^2
and
vf=vi + at

So....
x2=0 + 0t +1/2a(4) and x1=0 + 0t + 1/2a(36)
would the acceleration be -9.8 because of gravity? but I guess not because then I would get a negative answer.

4. Sep 16, 2008

### Redbelly98

Staff Emeritus
Looking good, you're almost there.

You're right that acceleration is not -9.8 m/s^2, but that's okay. Try taking the ratio x1/x2, like the problem asks, and use those two expressions you got.

5. Sep 17, 2008

### southernbelle

Okay, so it would be:

18a/2a.

So that would be 9a/a.

So vf=9a=a.
I am stuck :(

6. Sep 17, 2008

### Redbelly98

Staff Emeritus
Yes, so far so good. Can you simplify that fraction,

$$\frac{9a}{a}$$

Huh? No, that doesn't follow from anything, plus they are not asking for vf.

They are asking for the ratio of x1/x2, which you (correctly) got is 9a/a. You just need to simplify 9a/a, and then you're done.

7. Sep 17, 2008

### southernbelle

Ooohhh! So it would just be 9:1

?

Thanks so much!