Velocity-time graphs: finding acceleration from the gradient and distance travelled from the area under the line, including counting squares for a curved graph.

What this dot point is asking

Edexcel wants you to interpret velocity-time graphs: to find the acceleration from the gradient of the line and the distance travelled from the area under the line, including estimating the area under a curved graph by counting squares.

The two readings

This is the single most important fact for the graph: gradient gives acceleration, area gives distance. A common error is to mix this up with the distance-time graph, where the gradient gives speed and area has no meaning. A quick check is the units: gradient is velocity over time ($\text{m/s}^2$, an acceleration), and area is velocity times time ($\text{m}$, a distance).

Acceleration from the gradient

For a straight section, read a large change in velocity from the vertical axis and the matching change in time from the horizontal axis and divide. A horizontal line has zero gradient, so the acceleration is zero and the velocity is constant. A line sloping downwards has a negative gradient, meaning a deceleration.

Distance from the area

For a graph made of straight segments, divide the area into simple shapes and add their areas. For a curved graph, where no exact shape fits, Edexcel expects you to count the squares under the curve: work out the distance represented by one square from the axes, count the squares (counting a square as covered if more than half of it is under the line), and multiply. A neat shortcut for a trapezium-shaped section (a sloping line that does not start from zero) is to treat it as a rectangle plus a triangle, or to use the average of the two velocities multiplied by the time.

How Edexcel examines this

Velocity-time graphs are among the most heavily examined ideas in Topic 2, appearing on both tiers and often carrying four or more marks in a single structured question. A typical item describes a journey in stages (accelerate, cruise, decelerate) and asks you to calculate the acceleration of one stage from its gradient and the total distance from the whole area. The mark scheme rewards a clear method, so always label which quantity you are finding and show the triangle or rectangle areas you add. The most punished mistake is confusing this graph with a distance-time graph and reading the gradient when the question wants distance, or the area when it wants acceleration; the unit check (gradient gives $\text{m/s}^2$, area gives $\text{m}$) prevents this. Higher-tier papers may show a curved velocity-time graph and ask you to estimate the distance by counting squares, or to find the acceleration at a single instant by drawing a tangent, mirroring the tangent skill used on distance-time graphs. Reading the axes carefully matters: the vertical axis is velocity, so a horizontal line means a constant velocity and therefore zero acceleration, not a stationary object.

Try this

Q1. What does the area under a velocity-time graph represent? [1 mark]

• Cue. The distance travelled.

Q2. A velocity-time graph rises in a straight line from $0$ to $16\,\text{m/s}$ in $4\,\text{s}$. Calculate the acceleration. [2 marks]

• Cue. $a = \dfrac{16 - 0}{4} = 4\,\text{m/s}^2$.