Help - Uniform Acceleration Problem

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Homework Help Overview

The problem involves a boy on a skateboard accelerating uniformly down a hill, starting from rest, and focuses on determining the rate of acceleration based on the distance traveled during a specific time interval.

Discussion Character

  • Exploratory, Mathematical reasoning, Problem interpretation

Approaches and Questions Raised

  • Some participants discuss graphical methods, specifically finding the area under the graph to determine distance, while others express uncertainty about calculating velocities needed for this approach. An algebraic method is also suggested, involving equations of motion and substitution to solve for acceleration.

Discussion Status

The discussion is active, with participants exploring different methods to approach the problem. Some guidance has been offered regarding both graphical and algebraic techniques, but there is no explicit consensus on the best method to use.

Contextual Notes

Participants note the challenge of working with limited information, such as needing to find velocities at specific times and the implications of starting from rest.

petuniac
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Question is:

A boy on a skateboard accelerates uniformly down a hill, starting from rest. During the third one second interval from rest, he travels 7.5 m. What is the rate of acceleration of the skateboarder?

Answer is : 3.0 m/s^2

How do you do this problem??

Thanks!
 
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I do it graphically, find the area under the line between t=2 & t=3 and let it equal the distance. It is linear, so it is easily solved this way.
 
how do you graph when you don't have the slope of the line?
 
Hmm.. I think I'm missing something easy here. In order to calculate the area under the graph I need to know the velocity at time t=2 s and the velocity at time t = 3s. I'm not sure how to find this given only that v(initial) = 0 and d (between 2 and 3s) = 7.5 m...

Any help would be greatly appreciated.
 
OK, rather than do it graphically, let's try an algebraic method.

Take x(t) = 1/2 a t2, the general expression for distance as a function of 'constant' acceleration, i.e. accelerates uniformly.

Now let x(t=2s) = L = 1/2 a (2)2 (Eq 1), but that leaves two unknowns L and the acceleration a.

However, since we know that between 2s and 3s, the object move 7.5 m, then

let x(t=3s) = L + 7.5 m = . . . . (Eq 2)

then one had two unknowns and two equations. One can apply substitution, L from Eq 1 into L in Eq 2, and solve for 'a'.
 
Thanks! Got it now :)
 

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