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

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.