Constant acceleration of skateboarder

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Homework Help Overview

The problem involves a skateboarder coasting down a hill with constant acceleration, starting from rest. The skateboarder covers different distances over two time intervals, and the task is to find the ratio of these distances.

Discussion Character

  • Exploratory, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants discuss the initial assumption of a simple ratio based on time, questioning the implications of constant acceleration on distance covered.
  • Equations of motion under constant acceleration are referenced, and participants explore the correct application of these equations to find the distances.
  • There is a discussion about the nature of acceleration and its impact on the calculations, with some uncertainty about the value of acceleration.

Discussion Status

Participants are actively engaging with the problem, sharing their attempts and clarifying misunderstandings. Guidance has been provided to focus on the ratio of distances rather than velocity, and there is a productive exploration of the mathematical relationships involved.

Contextual Notes

There is an ongoing discussion about the correct interpretation of acceleration in the context of the problem, and participants are working through the implications of their assumptions.

southernbelle
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Homework Statement


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.


Homework Equations





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?
 
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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?
 
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.
 
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.
 
Okay, so it would be:

18a/2a.

So that would be 9a/a.

So vf=9a=a.
I am stuck :(
 
southernbelle said:
Okay, so it would be:

18a/2a.

So that would be 9a/a.

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

[tex] \frac{9a}{a}[/tex]

So vf=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.
 
Ooohhh! So it would just be 9:1

?

Thanks so much!
 

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