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A small car that we made in class was travelling and its time was recorded at 1,2,3, and 4 meters. The points in the form (seconds, meters) are as follows:

pt1 = (2.13,1)
pt2 = (3.48,2)
pt3 = (5.09,3)
pt4 = (6.75,4)

The question is to determine the distance traveled after 3.48 s, from the velocity-time graph.

I divided the graph into three parts and multiplied the time by the velocity for each section. enter image description here

A1 = (2.13)(0.46948) / 2 = 0.4999962 m
A2 = (3.48-2.13)(0.74074-0.46948) / 2 = (1.35)(0.27126) / 2 = 0.1831005 m
A3 = (0.46948)(1.35) = 0.633798 m

Total A = 0.4999962 m + 0.1831005 m + 0.633798 m = 1.3168947 m

I think this is wrong because in the point list, the distance traveled at 3.48 seconds is 2 m.

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marked as duplicate by Kyle Kanos, John Rennie, Jim, Brandon Enright, jinawee Mar 21 '14 at 22:38

This question was marked as an exact duplicate of an existing question.

  • $\begingroup$ Just looking at your graph, your calculation of areas is correct. So your graph must be wrong. In particular, your y-axis heights are incorrect. For example, the first triangle should reach a height of 0.938967, not 0.46948. $\endgroup$ – DumpsterDoofus Mar 21 '14 at 3:21
  • $\begingroup$ Reposting your question that was closed is not acceptable practice here. $\endgroup$ – Kyle Kanos Mar 21 '14 at 15:38
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Can I suggest a different way of doing this?

The first thing you should always do with experimental data is graph it to get an idea what it looks like. The graph of your data looks like:

Graph

The blue dots show the data points, and they obviously lie on a straight line (give or take experimental error). So I used linear regression to fit a straight line to the data and the pink line shows the fit. The equation of this line is:

$$ s = 0.605t - 0.111 \tag{1} $$

So the velocity is $0.6$ m/sec. The non-zero $y$ intercept is presumably due to experimental error in resording the data. You don't have to use a linear regression - you can draw a best fit straight line by eye then read the gradient off the graph.

Anyhow, you're asked for the distance at $3.48$ seconds, which seems a bit odd because that point is in your original data. Anyhow, to get the distance for any time $t$ just use the equation (1) that we got from from the fit. In this case we get:

$$ s = 0.605 \times 3.48 - 0.111 = 1.99 $$

And of course this is the correct answer because we already know the distance was $2$ m when the time was $3.48$ seconds.

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