Velocity-time graphs

These points form three straight sections. With real measurements, draw a line or smooth curve that represents the trend; small measurement variations should not automatically become sharp changes in motion.

Data for Velocity changes during an illustrative journey

Reading velocity and acceleration

The height of the graph tells you the velocity at that time. Its gradient, or slope, tells you how quickly velocity is changing: the acceleration.

From 0 to 4 s in the graph, velocity increases steadily. The straight upward-sloping line shows constant positive acceleration. From 4 to 6 s, the horizontal line shows constant velocity at 8 m/s, so acceleration is zero. From 6 to 8 s, velocity falls steadily to zero: the object decelerates.

A steeper line represents a greater magnitude of acceleration, provided the axis scales are the same. A straight sloping line has constant acceleration because equal time intervals produce equal changes in velocity. A curved line has changing acceleration because its gradient changes along the curve.

A horizontal line at zero velocity represents an object at rest. A horizontal line above zero represents an object moving at constant velocity.

Calculating acceleration from a gradient

Choose two well-separated points on the same straight section. A gradient triangle makes the vertical change in velocity and horizontal change in time clear. Divide the vertical change by the horizontal change:

$$a = \text{gradient} = \frac{\Delta v}{\Delta t}$$

On the first section of the graph, the points $(0,0)$ and $(4,8)$ give:

$$a = \frac{8-0}{4-0} = 2\text{ m/s}^2$$

This means velocity increases by 2 m/s each second.

On the final section, use $(6,8)$ and $(8,0)$:

$$a = \frac{0-8}{8-6} = -4\text{ m/s}^2$$

The negative gradient means velocity is decreasing. Here the object is moving in the positive direction and slowing down, so its deceleration has a magnitude of 4 m/s². The final section is steeper than the first and represents a faster change in velocity.

Finding distance from area — Higher Tier only

The area between a velocity-time line and the time axis gives displacement. When the object does not reverse direction and velocity is positive, this also gives the distance travelled.

The connection is easiest to see for constant velocity. Travelling at 8 m/s for 2 s covers 16 m. On the graph, this motion forms a rectangle with height 8 m/s and width 2 s. Its area is $8\times2=16$ m: velocity multiplied by time has units of distance.

For straight sections, divide the area into rectangles and triangles:

$$\text{Rectangle area} = \text{base}\times\text{height}$$ $$\text{Triangle area} = \frac{1}{2}\times\text{base}\times\text{height}$$

For the whole journey, the enclosed area consists of:

• 0–4 s: a triangle, $\frac{1}{2}\times4\times8=16$ m.
• 4–6 s: a rectangle, $2\times8=16$ m.
• 6–8 s: a triangle, $\frac{1}{2}\times2\times8=8$ m.

The total distance is $16+16+8=40$ m. All velocities are non-negative, so the displacement is also 40 m in the positive direction.

If a graph goes below the time axis, velocity is negative and the object is moving in the opposite direction. For displacement, areas below the axis count as negative. For total distance, add the magnitudes of all areas instead.

Estimating a curved area by counting squares — Higher Tier only

A curved graph cannot usually be divided exactly into a few triangles and rectangles. You can estimate its area using the graph's grid.

First calculate what one grid square represents, using its width on the time axis and its height on the velocity axis. Then count the complete squares between the curve and the time axis, combining partial squares to estimate additional whole squares. Multiply the estimated number of squares by the distance represented by each square.

For example, if each square spans 100 s horizontally and 5 m/s vertically, each represents $100\times5=500$ m. An estimated enclosed area of 17 squares would therefore represent $17\times500=8500$ m. This is an estimate because judging the partial squares introduces uncertainty.

Time on the horizontal axis; velocity on the vertical axis. Label units, choose even scales, plot time–velocity pairs and draw a line or curve representing the trend.

Area gives displacement — HT only

$$\text{Rectangle area}=bh,\qquad \text{Triangle area}=\frac{1}{2}bh$$

• Split straight-line graphs into shapes and add their areas.
• Curved graphs: count whole squares and combine partial squares.
• Distance per square = time width × velocity height.
• Area units: metres.
• Signed area gives displacement; add area magnitudes for total distance. These are equal for motion entirely in the positive direction.