# Comparing X1 to X2

1. Jun 16, 2008

### student 1

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 distance X1. On a second trip down the same hill starting from rest, he coasts for 2.00 seconds, covering a distance X2. Find the ratio of X1 to X2.

2. Relevant equations
V1= X1/6.00 s
V2= X2/2.00 s

3. The attempt at a solution
What is it talking about when it says find the ratio. I do not know what I need to find. I'm sure I'm suppose to calculate acceleration in somewhere, but I don't even know where to begin.

2. Jun 16, 2008

### Kurdt

Staff Emeritus
The ratio of x1 to x2 is simply x1/x2.

3. Jun 16, 2008

### student 1

Ok, so I'm assuming I need to find the distance that X1 and X2 equal. How do u determine that if you do not know the speed they traveled?

4. Jun 16, 2008

### Kurdt

Staff Emeritus
The questions says the skater started from rest and was subject to a constant acceleration. If you set up and equation for both distances in terms of these quantities and then divide them some terms will of course cancel out.

5. Jun 16, 2008

### HallsofIvy

Staff Emeritus
He didn't travel any specific speed- he had a constant acceleration. Assuming constant acceleration a, the first time he traveled for 6 seconds so he had an initial speed of 0 and final speed of 6a. The second time, he traveled for only 2 seconds so he had an initial speed of 0 and final speed of 2a.

Fortunately, with constant acceleration, the average speed is just the average of the two "end" speeds- the average speed is just the initial speed (0 in both cases) plus the final speed (6a and 2a) divided by 2.

(Actually, if you remember that the distance covered at constant acceleration depends upon t2, this is almost trivial! You can use that as a check.)

Last edited: Jun 16, 2008
6. Jun 16, 2008

### student 1

So basically, to find the ratio of the distance that is simply saying that You would divide the 6 seconds by 2 seconds and it would give you 3 therefore your distance for X1 would be 3 times greater than that of X2, due to them both accelerating at the same speed, is this correct?

7. Jun 16, 2008

### Kurdt

Staff Emeritus
No thats not correct. How would you calculate the distance normally. That is, what equation would you use?

8. Jun 16, 2008

### student 1

Xf=Xi+Vit+1/2(Acceleration)t^2

9. Jun 16, 2008

### Kurdt

Staff Emeritus
You know that the initial speed and distance are zero so both distances only depend on acceleration and time. Now divide x1 by x2.

10. Jun 16, 2008

### student 1

So it would be 6A/2A which would give you 3. Like before, but you cannot simply assume that it would be 6 divided by 2 right?

11. Jun 16, 2008

### Kurdt

Staff Emeritus
No. Where did you get 6a/2a from the equation you posted before? The acceleration term in that equation is multiplied by t2 and divided by 2.

12. Jun 16, 2008

### student 1

Oh, I'm sorry. I'm so confused with this it is not even funny! I understand the part you are talking about. So I got 9 for the ratio would that be right?

13. Jun 16, 2008

### Kurdt

Staff Emeritus
Yep. So its basically just the ratio of the times squared which is what HallsofIvy was alluding to before.

14. Jun 16, 2008

### student 1

THANK YOU SO MUCH!! I understood that it was simply this distance of x1 over x2, however I seem to try and make things way more hard than they are suppose to be! Thanks again!

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