use graphical methods to represent distance, displacement, speed, velocity and acceleration

Equations of Motion – Graphical Representation

In this lesson we explore how the fundamental kinematic quantities – distance, displacement, speed, velocity and acceleration – can be visualised using graphs. Understanding the shape and slope of these graphs allows you to extract numerical information without solving algebraic equations.

Key Definitions

• Distance ($s$): total length of the path travelled, always positive.
• Displacement ($\Delta x$): change in position, a vector quantity that can be positive or negative.
• Speed ($v$): magnitude of velocity, $v = \frac{ds}{dt}$, always positive.
• Velocity ($\vec v$): rate of change of displacement, $\vec v = \frac{d\vec x}{dt}$, can be positive or negative.
• Acceleration ($a$): rate of change of velocity, $a = \frac{dv}{dt}$, can be positive or negative.

Graph Types and Their Physical Meaning

1. Distance–time graph ($s$–$t$)
   • The gradient (slope) gives the speed: $v = \frac{ds}{dt}$.
   • A horizontal line ($v=0$) indicates the object is at rest.
   • A curved line indicates changing speed (i.e., acceleration).
2. Displacement–time graph ($x$–$t$)
   • The gradient gives the velocity: $\vec v = \frac{dx}{dt}$.
   • Positive slope → motion in the positive direction; negative slope → motion opposite to the chosen positive direction.
   • Zero slope → momentarily at rest.
3. Velocity–time graph ($v$–$t$)
   • The gradient gives the acceleration: $a = \frac{dv}{dt}$.
   • The area under the curve between two times gives the displacement: $\Delta x = \int v\,dt$.
   • A horizontal line indicates constant velocity (zero acceleration).
4. Acceleration–time graph ($a$–$t$)
   • The gradient gives the rate of change of acceleration (jerk), rarely needed at A‑Level.
   • The area under the curve gives the change in velocity: $\Delta v = \int a\,dt$.

Standard Equations of Motion (Constant Acceleration)

When acceleration $a$ is constant, the following equations relate the kinematic variables. They can be derived graphically from the shapes of the $v$–$t$ and $s$–$t$ graphs.

$$\begin{aligned} v &= u + at \\ s &= ut + \tfrac{1}{2}at^{2} \\ v^{2} &= u^{2} + 2as \end{aligned}$$

where $u$ is the initial velocity, $v$ the final velocity, $a$ the constant acceleration, $t$ the elapsed time and $s$ the displacement (or distance if motion is in a straight line without reversal).

Summary Table

Quantity	Symbol	SI Unit	Typical Graph	Interpretation of Slope / Area
Distance	$s$	metre (m)	$s$–$t$	Slope = speed $v$
Displacement	$\Delta x$	metre (m)	$x$–$t$	Slope = velocity $\vec v$
Speed	$v$	metre per second (m s⁻¹)	$v$–$t$	Horizontal line → constant speed
Velocity	$\vec v$	metre per second (m s⁻¹)	$v$–$t$	Slope = acceleration $a$; area = displacement
Acceleration	$a$	metre per second squared (m s⁻²)	$a$–$t$	Horizontal line → constant acceleration; area = change in velocity

Worked Example (Graphical Approach)

Consider a car that starts from rest, accelerates uniformly for 5 s, then moves at constant speed for another 5 s. The $v$–$t$ graph is a right‑angled triangle followed by a rectangle.

1. From the triangular part, the gradient gives the acceleration: $$a = \frac{\Delta v}{\Delta t} = \frac{20\ \text{m s}^{-1}}{5\ \text{s}} = 4\ \text{m s}^{-2}.$$
2. The area under the triangle (0–5 s) gives the displacement during acceleration: $$\Delta x_{1} = \tfrac{1}{2}\times 5\ \text{s}\times 20\ \text{m s}^{-1}=50\ \text{m}.$$
3. The area under the rectangle (5–10 s) gives the displacement at constant speed: $$\Delta x_{2}=20\ \text{m s}^{-1}\times5\ \text{s}=100\ \text{m}.$$
4. Total displacement after 10 s: $$\Delta x_{\text{total}} = 50\ \text{m}+100\ \text{m}=150\ \text{m}.$$

Practice Questions

1. From a given $s$–$t$ graph that is a parabola opening upwards, determine the expression for speed as a function of time and state whether the acceleration is constant.
2. A particle moves with a velocity described by $v = 3t^{2} - 12t + 9$ (units m s⁻¹, $t$ in seconds).
   • Find the acceleration as a function of time.
   • Calculate the displacement between $t = 1\ \text{s}$ and $t = 4\ \text{s}$ using the area under the $v$–$t$ curve.
3. Sketch an $a$–$t$ graph for a motion that starts from rest, accelerates uniformly for 2 s, then decelerates uniformly to rest over the next 3 s. Indicate the corresponding $v$–$t$ and $x$–$t$ shapes.

Key Take‑aways

• The slope of a graph gives the rate of change of the quantity plotted on the vertical axis.
• The area under a graph gives the accumulated change of the quantity on the vertical axis.
• For constant acceleration, the $v$–$t$ graph is a straight line and the $s$–$t$ graph is a parabola.
• Graphical analysis provides a quick check on algebraic calculations and helps visualise motion intuitively.