How Are Displacement, Velocity, and Acceleration Related Through Differentiation and Integration?

Velocity is the derivative of displacement, and acceleration is the derivative of velocity. Integration reverses these relationships: integrating acceleration gives velocity, while integrating velocity gives displacement.

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

The relationships

Let $s(t)$ represent displacement, $v(t)$ velocity, and $a(t)$ acceleration at time $t$.

Suppose

$$s(t)=t^3-6t^2+9t.$$

Differentiating gives

$$v(t)=3t^2-12t+9,$$

and differentiating again gives

$$a(t)=6t-12.$$

At $t=2$, $v(2)=-3$, so the particle is moving in the negative direction. Also, $a(2)=0$, which means the velocity has a stationary value at that instant; it does not mean the particle is stationary. A particle is stationary only when $$. Here, solving gives and . These times are the turning points of the displacement-time graph.

In reverse, integrating acceleration gives $v(t)=3t^2-12t+C$. An initial condition such as $v(0)=9$ is required to determine . Indefinite integration produces a family of functions, so the constant cannot be omitted.

The sign of velocity shows direction: $v(t)>0$ means displacement is increasing, while $v(t)<0$ means it is decreasing. However, $a(t)>0$ means velocity is increasing, not necessarily that the particle is moving forward.

Graphically, the gradient of a displacement-time graph is velocity, while the gradient of a velocity-time graph is acceleration. The signed area under a velocity-time graph gives displacement.

A common misconception is that integrating velocity always gives distance. It gives displacement:

$$\int_{t_1}^{t_2}v(t)\,dt.$$

Total distance requires speed:

$$\int_{t_1}^{t_2}|v(t)|\,dt.$$

Exam technique

Show the derivative or integral before giving a value. For total distance, find every time when $v(t)=0$, split the interval at sign changes, and integrate (|v(t)|). Do not report signed displacement as distance.