How can an integral find a distance traveled from a velocity function?

First write down the time interval and the velocity function, including units. Integrating velocity from the start time to the end time gives signed displacement: movement in one direction counts positive, and movement in the opposite direction counts negative. Total distance is different: it adds up all movement without allowing opposite directions to cancel. Mathematically, displacement is ∫v(t)dt, while total distance is ∫|v(t)|dt over the interval. [S1]

If velocity keeps the same sign throughout the interval, the magnitude of displacement equals total distance. If velocity changes sign, find where it is zero and account for each direction separately.

Assume a hypothetical object has velocity v(t)=t−2 metres per second for 0≤t≤4 seconds. This is an example for practicing the calculation, not a claim about a real trip.

The velocity is zero when t=2. It is negative from 0 to 2 seconds and positive from 2 to 4 seconds, so the object changes direction at t=2.

For signed displacement, integrate across the whole interval:

∫₀⁴(t−2)dt = [t²/2−2t]₀⁴ = 0 metres.

That zero means the two directions cancel in the net change in position; it does not mean the object stayed still. For total distance, split the calculation at the direction change and add the magnitudes of the two displacements. Each interval contributes 2 metres, so total distance is 2+2=4 metres. This follows the distinction between displacement and total distance described in the calculus source. [S1]

Find the times inside your interval where v(t)=0, then check whether the velocity changes sign there. A zero by itself does not necessarily mean the direction changes, so check the sign just before and after it if needed. Use the start and end times, together with any direction-change times, as the boundaries of your pieces.

For instance, in the example, t=2 divides [0,4] into [0,2] and [2,4]. The velocity is negative on the first piece and positive on the second. Integrate on each piece, take the absolute value of each result, and add them. Equivalently, integrate |v(t)| over the full interval. [S1]

Keep the time units consistent with the function. If velocity is in metres per second and time is in seconds, the integral has units of metres.

Label the interval and report both quantities separately if the question could mean either one. A compact working template is:

1. Interval: [start time, end time] = [___, ___] seconds.
2. Velocity: v(t)=___; solve v(t)=0 within the interval to find possible direction changes.
3. Displacement: ∫ from start to end of v(t)dt = ___ metres.
4. Total distance: split at direction changes; add the absolute values of the piecewise integrals = ___ metres.
5. Check: state the units and whether the answer is signed displacement or nonnegative total distance.

For the example, the entries are [0,4] seconds, v(t)=t−2, direction change at 2 seconds, displacement 0 metres, and total distance 4 metres. This method determines the mathematical quantities from the supplied velocity model; it does not establish whether that model accurately describes any real journey.