Topics

Distance–time graphs

Covers AS91171

What the axes mean

A distance–time graph (or position–time graph) has time on the horizontal axis and distance from a starting point on the vertical axis.
Each point on the line answers "where was it at that moment?"
The line is not the path the object took. A straight line does not mean the object moved in a straight line — it means the distance grew steadily.

The gradient is the velocity

$\text{gradient} = \frac{\text{rise}}{\text{run}} = \frac{\Delta d}{\Delta t} = v$

A steeper line means a faster object.
A horizontal line (zero gradient) means the object is stationary.
A negative gradient means the object is returning toward the starting point.

To find a velocity from the graph:

Choose two points on the line that are far apart and easy to read.
Subtract to get the rise ( $\Delta d$ ) and the run ( $\Delta t$ ).
Divide: $v = \Delta d / \Delta t$ .
Include the unit — m s⁻¹.

Reading the shape

Shape of line	What the object is doing
Horizontal	Stationary
Straight, sloping up	Constant velocity, moving away
Straight, sloping down	Constant velocity, returning
Steeper straight line	Same motion but faster
Curve getting steeper	Speeding up (accelerating)
Curve getting less steep	Slowing down

A curve on a distance–time graph always means the velocity is changing.

Instantaneous velocity from a curved graph

The gradient of a curve is different at every point, so a single "rise over run" across the whole curve gives only the average velocity.
To get the instantaneous velocity at one moment:
- Draw a tangent to the curve at that point — a straight line just touching the curve, with equal gaps above and below it either side.
- Extend the tangent a long way across the graph, so the triangle you read from is large.
- Find the gradient of that tangent.
Use a ruler — the specification says you must bring one, and this is one of the places you need it.

Worked ExampleReading velocity from a three-stage journey

A cyclist's distance–time graph shows: a straight line from $(0, 0)$ to $(20\text{ s}, 100\text{ m})$ ; a horizontal line from $(20\text{ s}, 100\text{ m})$ to $(50\text{ s}, 100\text{ m})$ ; then a straight line from $(50\text{ s}, 100\text{ m})$ back down to $(80\text{ s}, 0)$ . Describe each stage and find the velocity in each.

Stage 1 — $0$ to $20$ s

The line rises steadily, so the velocity is constant and away from the start:

$v_1 = \frac{\Delta d}{\Delta t} = \frac{100 - 0}{20 - 0} = 5.0 \text{ m s}^{-1} \text{ away from the start}$

Stage 2 — $20$ to $50$ s

The line is horizontal: the distance is not changing, so the cyclist is stationary for $30$ s.

$v_2 = \frac{0}{30} = 0 \text{ m s}^{-1}$

Stage 3 — $50$ to $80$ s

The line falls back to zero, so the cyclist is returning to the starting point:

$v_3 = \frac{0 - 100}{80 - 50} = \frac{-100}{30} = -3.3 \text{ m s}^{-1}$

The negative sign means the motion is back toward the start, at a speed of $3.3$ m s⁻¹.

$v_1 = 5.0 \text{ m s}^{-1} \text{ out}, \quad v_2 = 0, \quad v_3 = 3.3 \text{ m s}^{-1} \text{ back}$

Worked ExampleDistinguishing average from instantaneous velocity

A car's distance–time graph is a curve that starts flat at the origin and steepens, passing through $(10\text{ s}, 80\text{ m})$ . A tangent drawn at $t = 10$ s passes through $(4.0\text{ s}, 0)$ and $(12\text{ s}, 128\text{ m})$ . Find the average velocity over the first $10$ s and the instantaneous velocity at $t = 10$ s.

Step 1 — Average velocity uses the whole interval

Take the two end points of the interval, $(0, 0)$ and $(10, 80)$ :

$v_{\text{av}} = \frac{80 - 0}{10 - 0} = 8.0 \text{ m s}^{-1}$

Step 2 — Instantaneous velocity uses the tangent

The tangent touches the curve at $t = 10$ s, so its gradient is the velocity at that moment:

$v = \frac{128 - 0}{12 - 4.0} = \frac{128}{8.0} = 16 \text{ m s}^{-1}$

$v_{\text{av}} = 8.0 \text{ m s}^{-1}, \qquad v_{t = 10} = 16 \text{ m s}^{-1}$

Distance–time graphs

Covers AS91171

What the axes mean

A distance–time graph (or position–time graph) has time on the horizontal axis and distance from a starting point on the vertical axis.
Each point on the line answers "where was it at that moment?"
The line is not the path the object took. A straight line does not mean the object moved in a straight line — it means the distance grew steadily.

The gradient is the velocity

$\text{gradient} = \frac{\text{rise}}{\text{run}} = \frac{\Delta d}{\Delta t} = v$

A steeper line means a faster object.
A horizontal line (zero gradient) means the object is stationary.
A negative gradient means the object is returning toward the starting point.

To find a velocity from the graph:

Choose two points on the line that are far apart and easy to read.
Subtract to get the rise ( $\Delta d$ ) and the run ( $\Delta t$ ).
Divide: $v = \Delta d / \Delta t$ .
Include the unit — m s⁻¹.

Reading the shape

Shape of line	What the object is doing
Horizontal	Stationary
Straight, sloping up	Constant velocity, moving away
Straight, sloping down	Constant velocity, returning
Steeper straight line	Same motion but faster
Curve getting steeper	Speeding up (accelerating)
Curve getting less steep	Slowing down

A curve on a distance–time graph always means the velocity is changing.

Instantaneous velocity from a curved graph

The gradient of a curve is different at every point, so a single "rise over run" across the whole curve gives only the average velocity.
To get the instantaneous velocity at one moment:
- Draw a tangent to the curve at that point — a straight line just touching the curve, with equal gaps above and below it either side.
- Extend the tangent a long way across the graph, so the triangle you read from is large.
- Find the gradient of that tangent.
Use a ruler — the specification says you must bring one, and this is one of the places you need it.

Worked ExampleReading velocity from a three-stage journey

Stage 1 — $0$ to $20$ s

The line rises steadily, so the velocity is constant and away from the start:

$v_1 = \frac{\Delta d}{\Delta t} = \frac{100 - 0}{20 - 0} = 5.0 \text{ m s}^{-1} \text{ away from the start}$

Stage 2 — $20$ to $50$ s

The line is horizontal: the distance is not changing, so the cyclist is stationary for $30$ s.

$v_2 = \frac{0}{30} = 0 \text{ m s}^{-1}$

Stage 3 — $50$ to $80$ s

The line falls back to zero, so the cyclist is returning to the starting point:

$v_3 = \frac{0 - 100}{80 - 50} = \frac{-100}{30} = -3.3 \text{ m s}^{-1}$

The negative sign means the motion is back toward the start, at a speed of $3.3$ m s⁻¹.

$v_1 = 5.0 \text{ m s}^{-1} \text{ out}, \quad v_2 = 0, \quad v_3 = 3.3 \text{ m s}^{-1} \text{ back}$

Worked ExampleDistinguishing average from instantaneous velocity

Step 1 — Average velocity uses the whole interval

Take the two end points of the interval, $(0, 0)$ and $(10, 80)$ :

$v_{\text{av}} = \frac{80 - 0}{10 - 0} = 8.0 \text{ m s}^{-1}$

Step 2 — Instantaneous velocity uses the tangent

The tangent touches the curve at $t = 10$ s, so its gradient is the velocity at that moment:

$v = \frac{128 - 0}{12 - 4.0} = \frac{128}{8.0} = 16 \text{ m s}^{-1}$

$v_{\text{av}} = 8.0 \text{ m s}^{-1}, \qquad v_{t = 10} = 16 \text{ m s}^{-1}$

Distance–time graphs

What the axes mean

The gradient is the velocity

Reading the shape

Instantaneous velocity from a curved graph

Stage 1 — 000 to 202020 s

Stage 2 — 202020 to 505050 s

Stage 3 — 505050 to 808080 s

Step 1 — Average velocity uses the whole interval

Step 2 — Instantaneous velocity uses the tangent

Distance–time graphs

What the axes mean

The gradient is the velocity

Reading the shape

Instantaneous velocity from a curved graph

Stage 1 — 000 to 202020 s

Stage 2 — 202020 to 505050 s

Stage 3 — 505050 to 808080 s

Step 1 — Average velocity uses the whole interval

Step 2 — Instantaneous velocity uses the tangent

Stage 1 — $0$ to $20$ s

Stage 2 — $20$ to $50$ s

Stage 3 — $50$ to $80$ s

Stage 1 — $0$ to $20$ s

Stage 2 — $20$ to $50$ s

Stage 3 — $50$ to $80$ s