How Do You Find Total Distance Travelled Using Integration?

Integrate the absolute value of the velocity over the stated time interval. Equivalently, find when velocity changes sign, split the interval at those times, and add the positive distances from each section.

The Method

For an object with velocity $v(t)$ between times $t=a$ and $t=b$, the total distance travelled is

$$D=\int_{a}^{b}|v(t)|\,dt.$$

Velocity is the derivative of position, $v(t)=s'(t)$. A definite integral accumulates signed change: regions above the time axis contribute positively, while regions below it contribute negatively. Taking (|v(t)|) makes both types of motion contribute positively to path length.

Velocity may be negative when the object moves in the opposite direction, but distance cannot be negative. Therefore, integrating $v(t)$ without the absolute value gives displacement, not total distance.

On a velocity-time graph, distance is the sum of the areas between the graph and time axis, whether the graph lies above or below it.

If evaluating the absolute-value integral directly is inconvenient:

1. Solve $v(t)=0$ within the interval.
2. Determine the sign of $v(t)$ on each subinterval.
3. Reverse the sign of any integral where $v(t)<0$.
4. Add all contributions.

Worked Example

Suppose $v(t)=t-2$ for $0\leq t\leq5$. The direction changes when $t-2=0$, so . Since velocity is negative before this time and positive afterwards,

$$D=-\int_{0}^{2}(t-2)\,dt+\int_{2}^{5}(t-2)\,dt=2+4.5=6.5.$$

Here, the two sections contribute $2$ and $4.5$ units, and neither is cancelled. The displacement is only $\int_{0}^{5}(t-2)\,dt=2.5$.

IB Exam Technique

This is AHL 5.13 Kinematics, so it is HL-only. Write the integral setup before the GDC value, show every velocity zero used to split the interval, retain full precision, and include units. The common misconception is integrating velocity directly: that gives displacement unless velocity is non-negative throughout.