- #1

Bunny-chan

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## Homework Statement

I'm having some issues with the following exercise:

*The graph below represents the marking of a vehicle speedometer in funtion of time. Elaborate the corresponding graphs of acceleration and space traveled in function of time. What is the average acceleration of the vehicle between [itex]t = 0[/itex] and [itex]t = 1 min[/itex]? And between [itex]t=2min[/itex] and [itex]t = 3min[/itex]?*

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## Homework Equations

## The Attempt at a Solution

First I obtained all the equations of the straight lines, which are as follows:[tex]

v(t)= \begin{cases} 90t, \textrm{ if } 0 \leq t \leq 0.5 \\ 45, \textrm{ if } 0.5 \leq t \leq 2 \\ -90t+225, \textrm{ if } 2 \leq t \leq 2.5 \\ 0, \textrm{ if } 2.5 \leq t \leq 3 \\ 150t-450, \textrm{ if } 3 \leq t \leq 3.5 \\ 75, \textrm{ if } 3.5 \leq t \leq 4.5 \\ -150t+750, \textrm{ if } 4.5 \leq t \leq 5 \end{cases}[/tex]I know the acceleration graph data can be obtained by derivating the velocity functions and the displacement graph by integrating them, so:[tex]

a(t)= \begin{cases} 90, \textrm{ if } 0 \leq t \leq 0.5 \\ 0, \textrm{ if } 0.5 \leq t \leq 2 \\ -90, \textrm{ if } 2 \leq t \leq 2.5 \\ 0, \textrm{ if } 2.5 \leq t \leq 3 \\ 150, \textrm{ if } 3 \leq t \leq 3.5 \\ 0, \textrm{ if } 3.5 \leq t \leq 4.5 \\ -150, \textrm{ if } 4.5 \leq t \leq 5 \end{cases}[/tex][tex]x(t)= \begin{cases} 45t^2 + C, \textrm{ if } 0 \leq t \leq 0.5 \\ 45t + C, \textrm{ if } 0.5 \leq t \leq 2 \\ -90t^2+225t + C, \textrm{ if } 2 \leq t \leq 2.5 \\ 0, \textrm{ if } 2.5 \leq t \leq 3 \\ 75t^2-450t + C, \textrm{ if } 3 \leq t \leq 3.5 \\ 75t + C, \textrm{ if } 3.5 \leq t \leq 4.5 \\ -75t^2 +750t + C, \textrm{ if } 4.5 \leq t \leq 5 \end{cases}[/tex]And so I happily elaborated the graphs. And when I went to compare the results to the ones in my textbook, I realized there was no answer there. So I checked on the web and I found a document from an unknown source in which the question was solved, and I was faced with the following graphs:

*units*. It doesn't make sense to me how can the acceleration be around [itex]5000m~h^{-2}[/itex], for instance. I think it has something to do with the fact that the time is being measured in minutes, and perhaps I'm not converting it properly (I transformed minutes to seconds when finding the acceleration and position, and then converted it back to minutes. Am I being clear enough? D:) but I'm not so sure about that.

So, how do I convert velocity [itex](\frac{km}{h})[/itex] to position and acceleration? Any help to understand this would be greatly appreciated!

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