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I'm given the following information:

\begin{array}{|l|l|l|l|l|l|l|l|} \hline t & 0 & 0.5 & 1.0 & 1.5 & 2.0 & 2.5 & 3.0 \\ v(t) & 3.4 & 4.7 & 6.3 & 8.5 & 9.3 & 9.9 & 10.2 \\ \hline \end{array}

The table represents the velocity of a moving object at time t after passing a particular position.

I'm supposed to find how far the object moves from 0 to 3 seconds. Is this the correct way to do it?

$ \frac{3 - 0}{7} \cdot (3.4 + 4.7 + 6.3 + 8.5 + 9.3 + 9.9 + 10.2 ) $

Leaving me with the answer of: 22.41

EDIT

Left Hand Side:

$ \frac{1}{2}(3.4 + 4.7 + 6.3 + 8.5 + 9.3 + 9.9) $

21.05

Right Hand Side:

$ \frac{1}{2}(4.7 + 6.3 + 8.5 +9.3 +9.9 +10.2) $

24.45

Average:

22.75

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    I think I fixed the TeX you were going after, but please make sure that what I produced is what you wanted.2012-12-10
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    @mixedmath Thank you, that is exactly what I was trying to do.2012-12-10

1 Answers 1

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You can't determine how far the object moves because you don't know its velocity except at isolated points. You can get an estimate of how far it may have moved if its velocity didn't change too abruptly, by applying the trapezoidal rule. This would involve replacing the denominator $7$ by $6$ and weighting the first and last velocity values by $1/2$.

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    Does my edit above have the right idea?2012-12-10
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    @StrugglingWithMath: That's an interesting way of looking at the trapezoidal rule; it hadn't occurred to me before :-). It yields the same result, but you might also want to look at the more conventional interpretation as the sum of the areas of trapezoids (whence the name).2012-12-10
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    I will definitely look into it (as I'll be taking Calc II in the near future and from what I hear it goes into greater depths dealing with area). I was just wanting to make sure I understand the general idea of it prior to my test :) Thanks for the help!2012-12-10