Comparing X1 to X2

Homework Statement

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.

V1= X1/6.00 s
V2= X2/2.00 s

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.

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

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?

Kurdt
Staff Emeritus
Gold Member
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.

HallsofIvy
Homework Helper
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?

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.)

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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?

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

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

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

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?

Kurdt
Staff Emeritus