Motion along a straight line

Kinematics is a branch of applied mathematics that describes motion using position, velocity and acceleration without asking what force caused the motion. Here, we model one-dimensional motion. The object travels along a straight line, with every measurement taken relative to a fixed origin.

Displacement is a signed distance from a fixed origin that states where an object is relative to that origin. It is usually written as $s(t)$, where $s$ is displacement from the origin ($\text{m}$) and $t$ is time ($\text{s}$). The sign matters: $s(t)=3$ and $s(t)=-3$ are the same distance from the origin, but on opposite sides.

Distance travelled is a non-negative accumulated length of path that records how far the object has actually moved. If the object turns around, this won’t be the same as its displacement. For example, a runner who goes $10\ \text{m}$ east and then $10\ \text{m}$ west has displacement $0\ \text{m}$ but distance travelled $20\ \text{m}$.

Velocity is the rate of change of displacement with respect to time, including direction. We write

When velocity is positive, displacement is increasing. When it is negative, displacement is decreasing. Speed is the magnitude of velocity, a non-negative rate at which distance is being covered, so speed is $|v|$.

Acceleration is the rate of change of velocity with respect to time. We write

A positive acceleration means velocity is increasing, while a negative acceleration means velocity is decreasing. Be careful with the wording here. If an object is already moving in the negative direction, a negative acceleration can make it speed up.

Keep the calculus ladder clear: differentiate displacement to get velocity, then differentiate velocity to get acceleration. Going the other way, integrate acceleration to get velocity and integrate velocity to get displacement.

Differentiating a displacement model

When a question gives $s(t)$, differentiate it to find velocity and acceleration. Suppose, for example,

$$s(t)=6+4t-t^2$$

Then

$$v(t)=\frac{ds}{dt}=4-2t$$

and

$$a(t)=\frac{dv}{dt}=-2$$

The initial displacement is $s(0)$, and the initial velocity is $v(0)$. Unless the question clearly gives another starting time, "initial" means at $t=0$.

Integrating acceleration or velocity

If acceleration is given, integrate once for velocity and once more for displacement. Don’t ignore the constants of integration: they bring the initial conditions into the model.

For instance, if

$$a(t)=3\sin t$$

then

$$v(t)=\int 3\sin t\,dt=-3\cos t+C$$

where $C$ is a constant of integration with the same units as velocity ($\text{m s}^{-1}$). If the initial velocity is $2\ \text{m s}^{-1}$, then $v(0)=2$, so

$$2=-3\cos 0+C=-3+C$$

which gives $C=5$. Hence

$$v(t)=-3\cos t+5$$

Next, integrate velocity to find displacement:

$$s(t)=\int (-3\cos t+5)\,dt=-3\sin t+5t+D$$

where $D$ is a constant of integration with the same units as displacement ($\text{m}$). If the initial displacement is $1\ \text{m}$, then $s(0)=1$. This gives $D=1$, so

$$s(t)=-3\sin t+5t+1$$

Use a different letter for each constant. It’s a small habit that prevents a surprising number of algebra errors.

Displacement from a velocity-time graph

Between time $t_1$ and time $t_2$, displacement equals the signed area under the velocity-time graph:

Area above the time-axis is positive, while area below it is negative. The two can cancel, so displacement may be zero even though the object has moved.

For the velocity function $v(t)=\sin(2t)$ over the interval from $0$ to $\pi$, the positive and negative signed areas cancel. The displacement is therefore $0\ \text{m}$. The object did not stand still; it finished where it started.

Total distance travelled

To find the distance travelled between $t_1$ and $t_2$, integrate speed rather than velocity:

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

The absolute value is essential. It reflects every part of the velocity graph below the time-axis above it before the area is calculated. As a result, distance travelled can never be negative.

For an analytical calculation, begin by finding any times inside the interval when $v(t)=0$, since these are the possible turning times. Split the integral at those values and make every contribution positive. If technology is allowed, enter the absolute value integral directly when the question asks for total distance travelled.

Use the following interpretation:

• $v(t)>0$: displacement is increasing, so the object is moving in the positive direction.
• $v(t)<0$: displacement is decreasing, so the object is moving in the negative direction.
• $v(t)=0$: the object is instantaneously at rest; it may or may not be changing direction.

The modelling point

This topic shows the link between calculus and physics very clearly. Derivatives describe rates of change, while integrals accumulate change over time. The same mathematical structure can represent a falling object, a lift moving between floors, a vehicle travelling along a straight road or a particle moving on a line.

Why did kinematics become part of a core mathematics course? The historical development of calculus was strongly shaped by astronomy, mechanics and European scientific traditions, although the idea of describing change extends far beyond any one culture. Mathematics is often presented as culture-free. Yet people decide which examples to choose, which applications to value and even where to place the boundary between "mathematics" and "physics". The model may be abstract, but the choice of model is never completely neutral.