What's the Difference Between Distance and Displacement in a Kinematics Problem?

Displacement is the signed change in a particle's position, whereas distance is the total length of the path travelled. Distance is always non-negative; displacement may be positive, negative, or zero.

This is part of IB Mathematics: Analysis and Approaches SL 5.9: Kinematics problems and may be tested on Paper 1 or Paper 2.

The Key Difference

Velocity is the derivative of displacement: $v(t)=\dfrac{ds}{dt}$. Therefore, integrating velocity gives the net change in position.

If a position function $s(t)$ is given instead, displacement is $s(t_2)-s(t_1)$. To find distance, first differentiate to obtain , then locate every zero of velocity inside the stated interval.

For distance, any section where $v(t)<0$ must still contribute positively. You should find the times when $v(t)=0$, identify any changes of direction, and split the integral at those times.

Graphically, displacement is the signed area between the velocity graph and the time axis; distance is the sum of the magnitudes of those areas.

Worked Example

Suppose $v(t)=2t-4$ for $0\leq t\leq4$.

The displacement is

$$\int_{0}^{4}(2t-4)\,dt=\left[t^2-4t\right]_{0}^{4}=0.$$

However, $v(t)=0$ at $t=2$. The particle moves in the negative direction before $t=2$ and in the positive direction afterwards. Hence, the distance is

$$-\int_{0}^{2}(2t-4)\,dt+\int_{2}^{4}(2t-4)\,dt=4+4=8.$$

The particle travels 8 units but returns to its starting position, so its displacement is 0. This illustrates cancellation: equal movements in opposite directions produce zero net displacement, but both count toward distance.

A common misconception is that distance equals the absolute value of displacement. This is only true when the particle does not change direction.

Exam Technique

For a "find the total distance" question, show the equation $v(t)=0$, split the interval at every valid turning time, and write the correct integral expression before evaluating it. Do not integrate velocity over the whole interval without accounting for sign changes. Include units in the final answer.