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:

• determine the average (or constant) speed of an object by finding the gradient (slope) of a straight‑line section of a distance‑time graph,
• interpret what different shapes of a distance‑time graph signify,
• distinguish between speed (scalar) and velocity (vector), and
• link distance‑time and speed‑time graphs using the concepts of gradient and area.

Key Concepts

• Distance–time graph: vertical axis = distance (m), horizontal axis = time (s).
• Gradient (slope): \(\displaystyle \text{gradient}=\frac{\Delta d}{\Delta t}\) – the definition of average speed for the chosen interval.
• Uniform motion: a straight‑line segment → constant speed (gradient is the same at every point on that segment).
• Non‑uniform motion: a curved line → gradient changes; the gradient at a single point (tangent) gives the instantaneous speed.
• Velocity: a vector quantity; its magnitude is the speed and its direction is indicated by the sign of the gradient (positive = motion away from the origin, negative = motion back towards the origin).
• Acceleration (from a speed‑time graph): the gradient of a straight‑line segment of a speed‑time graph.

Qualitative Interpretation of a Distance‑Time Graph

From the shape you can tell:

Shape	What it Means
Horizontal line	Object at rest (gradient = 0)
Straight line, non‑zero gradient	Constant speed (uniform motion)
Curve that becomes steeper	Accelerating (gradient increasing)
Curve that becomes less steep	Decelerating (gradient decreasing)

Mathematical Relationships

Quantity	Formula	When to Use
Speed (gradient of distance‑time)	\(v=\dfrac{\Delta d}{\Delta t}\)	Straight‑line segment of a distance‑time graph
Average speed (overall)	\(v_{\text{avg}}=\dfrac{\text{total distance}}{\text{total time}}\)	Any motion (including curved sections)
Acceleration (gradient of speed‑time)	\(a=\dfrac{\Delta v}{\Delta t}\)	Straight‑line segment of a speed‑time graph
Distance from speed‑time graph	\(\text{distance}= \text{area under the speed‑time curve}\)	Any speed‑time graph (rectangle, triangle, trapezoid)

Link Between Distance‑Time and Speed‑Time Graphs

• The gradient of a distance‑time graph gives the object's speed (or the magnitude of its velocity).
• The area under a speed‑time graph gives the distance travelled.
• Conversely, the gradient of a speed‑time graph gives the object's acceleration.

Area‑under‑curve example (speed‑time)

Step‑by‑Step Procedure (Gradient Method)

1. Identify a straight‑line segment where the motion is uniform.
2. Choose two points on that segment (preferably the end points) and read their coordinates \((t_{1},d_{1})\) and \((t_{2},d_{2})\).
3. Calculate the change in distance: \(\Delta d = d_{2}-d_{1}\).
4. Calculate the change in time: \(\Delta t = t_{2}-t_{1}\).
5. Compute the speed: \(\displaystyle v = \frac{\Delta d}{\Delta t}\).
6. Record the answer with the correct units (m s\(^{-1}\)) and to the appropriate number of significant figures (see the reminder below).

Significant‑figure reminder

• When multiplying or dividing, keep as many decimal places as the measurement with the fewest decimal places.
• When adding or subtracting, keep as many decimal places as the measurement with the fewest decimal places.
• Express the final speed to the same number of significant figures (or decimal places) as the least‑precise measurement used in the calculation.

How to Sketch a Distance‑Time Graph from Data

1. Choose convenient scales (e.g., 1 cm = 2 s on the time axis, 1 cm = 5 m on the distance axis).
2. Plot each data point \((t,d)\) accurately.
3. If the motion between two points is uniform, join them with a straight line; label the line (e.g., “uniform motion”).
4. Label the axes with quantity and unit, and include a title.

Worked Example 1 – Straight‑Line Segment

Extracted data from a distance‑time graph:

Point	Time \(t\) (s)	Distance \(d\) (m)
A	2	4
B	6	20

Calculate the speed between A and B.

Solution:

\[ \Delta d = 20\;\text{m} - 4\;\text{m} = 16\;\text{m} \] \[ \Delta t = 6\;\text{s} - 2\;\text{s} = 4\;\text{s} \] \[ v = \frac{16\;\text{m}}{4\;\text{s}} = 4.0\;\text{m s}^{-1} \]

The gradient of the straight‑line segment AB is \(4.0\;\text{m s}^{-1}\); therefore the object moves at a constant speed of \(4.0\;\text{m s}^{-1}\) during this interval.

Worked Example 2 – Curved Distance‑Time Graph (Instantaneous Speed)

Consider the curve below, which represents a car that is accelerating.

To estimate the instantaneous speed at \(t = 3\;\text{s}\) (point C):

1. Draw a tangent that just touches the curve at C.
2. Read two points on the tangent, e.g. \((2.8\;\text{s}, 5.6\;\text{m})\) and \((3.2\;\text{s}, 7.2\;\text{m})\).
3. Calculate \(\Delta d = 7.2 - 5.6 = 1.6\;\text{m}\) and \(\Delta t = 3.2 - 2.8 = 0.4\;\text{s}\).
4. Instantaneous speed \(v_{\text{inst}} = \dfrac{1.6}{0.4}=4.0\;\text{m s}^{-1}\).

This demonstrates how a changing gradient yields varying instantaneous speeds.

Common Mistakes (and How to Avoid Them)

• Forgetting to divide – always use \(\Delta d / \Delta t\), not just \(\Delta d\).
• Mixing units – ensure distance and time are in the same system (m and s) before calculating.
• Applying a single gradient to a curved portion – gives only an average speed; use a tangent for instantaneous speed.
• Omitting units or using inconsistent units – write the answer with m s\(^{-1}\) (or km h\(^{-1}\) if that was the original unit).
• Incorrect significant figures – follow the rules in the “Significant‑figure reminder”.

Quick‑Check Checklist (before you submit)

• Did I use \(\Delta d / \Delta t\) for the gradient?
• Are the two points taken from a straight‑line (uniform) segment?
• Are the units m and s, giving a speed in m s\(^{-1}\) (or the required unit)?
• Is the answer reported to the correct number of significant figures?
• If the graph is curved, did I use a tangent for 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 \((1\;\text{s}, 3\;\text{m})\) and \((5\;\text{s}, 15\;\text{m})\). 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.
4. Sketch a distance‑time graph for an object that starts from rest, accelerates uniformly for 4 s to reach a speed of 6 m s\(^{-1}\), then moves at that constant speed for a further 6 s. (Mark the scales you would use.)
5. Given a speed‑time graph where the speed is constant at 5 m s\(^{-1}\) for 10 s, calculate the distance travelled using the “area under the curve” method.
6. From a speed‑time graph that shows a straight line increasing from 0 to 4 m s\(^{-1}\) over 5 s, find the acceleration.

Cross‑Reference Box

See also:

• 1.5 Forces – how net force relates to acceleration (Newton’s 2nd law).
• 1.3 Velocity – vector nature of motion, direction of motion, and how to represent velocity on graphs.
• 1.4 Acceleration – gradient of speed‑time graphs and its link to forces.

Summary

The gradient of a straight‑line section of a distance‑time graph directly yields the constant speed (or the magnitude of the velocity) of the object during that interval:
\(\displaystyle v = \frac{\Delta d}{\Delta t}\).
For non‑uniform motion, a tangent provides the instantaneous speed. The same graphical ideas link distance‑time, speed‑time, and acceleration‑time graphs through gradients and areas, fulfilling the core requirements of Cambridge IGCSE Physics 0625 – Motion.