Calculate speed from the gradient of a straightline section of a distance-time graph

1.2 Motion – Calculating Speed from a Distance‑Time Graph

Learning Objective

Students will be able to calculate the speed of an object by determining the gradient (slope) of a straight‑line section of a distance‑time graph.

Key Concepts

• Distance–time graph: vertical axis = distance (m), horizontal axis = time (s).
• Gradient of a straight line = \$\displaystyle \frac{\Delta d}{\Delta t}\$, which is the definition of average speed.
• For a uniform motion (constant speed) the graph is a straight line; the gradient is the constant speed.

Mathematical Relationship

The speed \$v\$ is given by

\$v = \frac{\Delta d}{\Delta t}\$

where \$\Delta d\$ is the change in distance and \$\Delta t\$ is the corresponding change in time.

Step‑by‑Step Procedure

1. Identify a straight‑line segment on the distance‑time graph where the motion is uniform.
2. Choose two points on that segment (preferably at the ends) and read their coordinates \$(t1, d1)\$ and \$(t2, d2)\$.
3. Calculate the change in distance: \$\Delta d = d2 - d1\$.
4. Calculate the change in time: \$\Delta t = t2 - t1\$.
5. Compute the speed: \$v = \dfrac{\Delta d}{\Delta t}\$.
6. Check units: if distance is in metres and time in seconds, speed will be in \$\text{m s}^{-1}\$.

Worked Example

Consider the following distance‑time data extracted from a graph:

Calculate the speed between points A and B.

Solution:

\$\Delta d = 20\ \text{m} - 4\ \text{m} = 16\ \text{m}\$

\$\Delta t = 6\ \text{s} - 2\ \text{s} = 4\ \text{s}\$

\$v = \frac{16\ \text{m}}{4\ \text{s}} = 4\ \text{m s}^{-1}\$

The gradient of the straight‑line segment AB is \$4\ \text{m s}^{-1}\$, therefore the object moves at a constant speed of \$4\ \text{m s}^{-1}\$ during this interval.

Common Mistakes

• Using the vertical distance between two points on the graph as the speed directly, without dividing by the horizontal time interval.
• Reading the axes incorrectly (e.g., mixing up metres and kilometres).
• For curved sections, the gradient changes; using a single gradient there gives only an average speed, not the instantaneous speed.

Practice Questions

1. A car travels uniformly for 8 s covering a distance of 32 m. Using the distance‑time graph method, find its speed.
2. On a distance‑time graph, a straight line passes through the points (1 s, 3 m) and (5 s, 15 m). Determine the speed and state the units.
3. Explain why the gradient of a curved portion of a distance‑time graph cannot be used to find the instantaneous speed.

Suggested Diagram

Summary

The gradient of a straight‑line section of a distance‑time graph directly gives the constant speed of an object during that interval. By applying the formula \$v = \Delta d / \Delta t\$, students can convert graphical information into quantitative speed values, a fundamental skill in IGCSE Physics.