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If the values in the table were graphed what would be the slope of the line? x:time(s) 6, 13, 15, 20 y:distance (m) 1396, 3020, 3484, 4644
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232
slope is dy/dx
How did you get that?
\[m={y_2-y_1\over x_2-x_1}={3020-1396\over13-6}=232\]
m1 above\[m_2={y_3-y_2\over x_3-x_2}={3484-1320 \over15-13}=1082\]continue this pattern for each segment and you will have the slope for various intervals. the average slope across the whole data set is\[{y_4-y_1\over x_4-x_1}={4644-20\over20-6}={2312\over7}\]you can also find the average slope in each interval with \[m_{avg}={m_1+m_2+m_3\over3}={232+1082+{1162\over5}\over3}=2399.2\]Another method is to find the best fit curve, but this is far more complex than what I have given.
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