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)
- Identify a straight‑line segment where the motion is uniform.
- Choose two points on that segment (preferably the end points) and read their coordinates \((t{1},d{1})\) and \((t{2},d{2})\).
- Calculate the change in distance: \(\Delta d = d{2}-d{1}\).
- Calculate the change in time: \(\Delta t = t{2}-t{1}\).
- Compute the speed: \(\displaystyle v = \frac{\Delta d}{\Delta t}\).
- 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
- Choose convenient scales (e.g., 1 cm = 2 s on the time axis, 1 cm = 5 m on the distance axis).
- Plot each data point \((t,d)\) accurately.
- If the motion between two points is uniform, join them with a straight line; label the line (e.g., “uniform motion”).
- 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):
- Draw a tangent that just touches the curve at C.
- Read two points on the tangent, e.g. \((2.8\;\text{s}, 5.6\;\text{m})\) and \((3.2\;\text{s}, 7.2\;\text{m})\).
- Calculate \(\Delta d = 7.2 - 5.6 = 1.6\;\text{m}\) and \(\Delta t = 3.2 - 2.8 = 0.4\;\text{s}\).
- 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
- A car travels uniformly for 8 s covering a distance of 32 m. Using the distance‑time graph method, find its speed.
- 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.
- Explain why the gradient of a curved portion of a distance‑time graph cannot be used to find the instantaneous speed.
- 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.)
- 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.
- 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.