2.1.6Forces

Area Under Velocity-Time Graphs

5 min read 8 parts Lesson 7 of 33 1 quiz

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Section 1 of 8Why the area gives distance

For constant speed, distance = speed × time.

Area of the rectangle

On a velocity-time graph, speed is the height and time is the width of a rectangle, so the product of the two is the area of the rectangle.

The same idea works when the speed changes: divide the journey into many short intervals in each of which the speed is nearly constant, add the areas of the thin strips,

and the total is the area under the line.

distance travelled = area under the velocity-time graph

The units come from the axes

the area has units of m/s × s = m.
If the velocity is in km/h and the time is in hours, the distance is in km.

Section 2 of 8Common shapes

Split the area under the graph into simple shapes, find the area of each and add them:

Rectangle

(constant velocity): area = velocity × time.

Triangle

(steady acceleration from rest, or steady braking to rest): area = ½ × base × height = ½ × time × velocity.

Trapezium

(steady change from one non-zero velocity to another): area = ½ × (sum of the parallel sides) × width = ½ (u + v) t.

This is the same as the average velocity multiplied by time.

For a graph that rises from rest, stays constant and falls back to rest, the shape is a trapezium, or you can treat it as a triangle, a rectangle and another triangle. Either way gives the same answer. Always use the correct times along the horizontal axis for the base of each shape, and not the total time.

Section 3 of 8Curves and counting squares

If the graph is curved, you cannot use the simple formulas. Estimate the area by counting squares:

Find the area represented by one square (the height value × the width value of a square).

For example, if one square is 2 m/s high and 1 s wide, then 1 square = 2 m.

Count the whole squares under the line.

Count the squares which are more than half covered as whole ones, and ignore those less than half covered. (Alternatively, add up the part squares.)

Multiply the total number of squares by the area of one square.

The answer is an estimate; the more precise the grid, the better the estimate.

Section 4 of 8Negative velocity and displacement

If the graph goes below the time axis, the velocity is negative, which means the object is moving back the other way.

Displacement

The area below the axis counts as negative displacement.

Displacement

The displacement is the net change in position, found by subtracting the area below from the area above.

Distance

The total distance travelled is the sum of both areas, both counted as positive.

This lesson links to the equations for uniform acceleration (1.1.7): the area of the trapezium ½(u + v)t is the origin of the equation s = ½(u + v)t.

Section 5 of 8Working through a typical question

A bus accelerates uniformly from rest to 10 m/s in 5.0 s, travels at 10 m/s for 20 s, then brakes to rest uniformly in 10 s.

To find the total distance, sketch the graph: a trapezium with parallel sides 35 s (the total time) along the bottom and 20 s along the top,

and a height of 10 m/s.

Area = ½ (35 + 20) × 10 = 275 m.

Check by splitting: ½ × 5.0 × 10 = 25 m, 10 × 20 = 200 m, ½ × 10 × 10 = 50 m; the sum is 275 m, so the two methods agree.

The average speed for the journey is 275 ÷ 35 = 7.9 m/s, which is lower than the cruising speed because of the slower start and end.

When you use the area method, show the shapes you use, give each area with units, and add them to find the total. Doing this carefully earns method marks even if there is an arithmetic slip.

Section 6 of 8Worked examples

A car moves at a constant 15 m/s for 20 s. Find the distance travelled from the graph.
A train accelerates uniformly from rest to 24 m/s in 30 s, travels at constant speed for 60 s, then brakes uniformly to rest in 20 s. Find the total distance.
A velocity-time graph is a straight line from 4.0 m/s at 0 s to 10 m/s at 5.0 s. Find the distance travelled, and the average velocity.

Section 7 of 8Key terms

Term	Meaning
Area under graph	Equal to distance travelled on a velocity-time graph.
Trapezium	Four-sided shape with one pair of parallel sides; area = ½(a + b)h.
Displacement	Change in position in a given direction; a vector.
Average velocity	Total displacement divided by total time.
Counting squares	A method for estimating the area under a curve.

Section 8 of 8Exam technique

Using the whole time as the base for each part of the graph.

Forgetting the ½ when finding the area of a triangle.

Taking the gradient instead of the area to find distance.

Getting the units wrong. Distance is in m, not m/s.

Counting the area below the axis as positive when asked for displacement.

Physics 0625 & 9702 SUVAT Mastery Graph Interpretation

Constant Acceleration Equations (SUVAT) & Slopes

v = u + at | s = ut + ½at² | v² = u² + 2as | s = ½(u + v)t

Displacement-Time Graph: Gradient = Velocity (curve = accelerating).
Velocity-Time Graph: Gradient = Acceleration; Area under curve = Distance travelled / Displacement.
Free Fall in vacuum: a = g ≈ 9.81 m/s² downwards (air resistance negligible).