Net Change vs Total Distance

Displacement and Distance from Velocity

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In calculus, net change and total distance both come from the area under a velocity versus time graph, but they answer different physical questions. Net change tells how far the position has shifted from the starting point, including direction. Total distance tells how much ground was actually covered, no matter which way the object moved. This distinction matters whenever motion changes direction.

If velocity stays positive, net change and total distance are the same. When velocity becomes negative, the signed area below the time axis subtracts from the signed area above it, so net change can be smaller than the total distance. Total distance is found by adding the magnitudes of all areas, which is equivalent to integrating the absolute value of velocity. A graph that crosses the time axis is the clearest way to see the difference.

Key Facts

Net change in position over $[a, b]$ is $\Delta x = \int_a^b v(t)\,dt$ .
Total distance traveled over $[a, b]$ is $D = \int_a^b |v(t)|\,dt$ .
Area above the time axis contributes positively to the integral of $v(t)\,dt$ .
Area below the time axis contributes negatively to the integral of $v(t)\,dt$ .
If v(t) does not change sign on [a, b], then net change and total distance have the same magnitude.
To compute total distance from a velocity graph, split the interval at times where $v(t) = 0$ and add absolute areas.

Vocabulary

Net change: The overall change in a quantity over an interval, found by adding positive and negative contributions together.
Total distance: The full amount of motion traveled, found by adding all movement as positive amounts.
Velocity: A rate of change of position that includes both speed and direction.
Signed area: Area counted as positive above the axis and negative below the axis.
Absolute value: The distance of a number from zero, so it is always nonnegative.

Common Mistakes to Avoid

Using $\int_a^b v(t)\,dt$ for total distance, which is wrong because negative velocity subtracts instead of adding to the amount traveled.
Ignoring where the velocity graph crosses the time axis, which is wrong because sign changes determine where you must split the integral.
Treating negative velocity as negative distance, which is wrong because distance is always nonnegative and measures amount of travel, not direction.
Confusing position with velocity on the graph, which is wrong because the area under a velocity graph gives change in position, not the position itself.

Practice Questions

1 A particle has velocity $v(t) = 4 - 2t$ for $0 \leq t \leq 4$ . Find the net change in position and the total distance traveled.
2 A velocity graph forms a triangle above the time axis from $t = 0$ to $t = 3$ with height $6$ , and a triangle below the time axis from $t = 3$ to $t = 5$ with height $4$ . Find the net change in position and the total distance traveled.
3 Explain why an object can have zero net change in position over a time interval but still have a positive total distance traveled.

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