Motion Graphs: Distance–Time and Velocity–Time Graphs

Pictures of Motion

A single graph can tell the whole story of a journey: when a bus sped up, how long it waited at a stop, how fast it travelled and how far it went. That is why examiners love motion graphs. In WAEC, NECO and JAMB physics you are regularly given a distance–time or velocity–time graph and asked to read off a speed, calculate an acceleration or find the distance travelled. This tutorial explains how to read and draw motion graphs: what the gradient and the area under each graph mean, how to recognise rest, uniform velocity and uniform acceleration, and how to solve typical exam questions step by step. It builds on the tutorial on the equations of motion.

Key Quantities

Distance is the total length of path travelled. It is a scalar, measured in metres (m).
Displacement is the distance moved in a stated direction from the starting point. It is a vector.
Speed is the rate of change of distance, and velocity is the rate of change of displacement. Both are measured in m/s.
Acceleration is the rate of change of velocity, a = (v − u)/t, measured in m/s². A negative acceleration is called a deceleration or retardation.

Every motion graph is built from these quantities, and two ideas unlock all of them: the gradient (slope) of a graph and the area under the graph.

Distance–Time Graphs

A distance–time (or displacement–time) graph plots distance on the vertical axis against time on the horizontal axis.

The gradient gives the speed

Gradient = change in distance ÷ change in time = speed. So the steeper the line, the faster the object is moving.

A horizontal line means the distance is not changing: the object is at rest.
A straight sloping line means equal distances in equal times: uniform (constant) speed.
A curve getting steeper means the speed is increasing: the object is accelerating.
A curve getting flatter means the object is slowing down.

For a curved graph, the speed at an instant is the gradient of the tangent drawn to the curve at that point.

Worked example 1

A cyclist rides 300 m away from home in 60 s, stops for 40 s, then rides back home in 50 s. Describe the displacement–time graph and find (a) the speed on each stage, (b) the average speed and (c) the average velocity for the whole trip.

The graph rises in a straight line from (0, 0) to (60 s, 300 m), stays horizontal at 300 m until 100 s, then falls in a straight line to 0 m at 150 s.

(a) Stage 1: 300 ÷ 60 = 5 m/s. Stage 2: horizontal, so 0 m/s. Stage 3: 300 ÷ 50 = 6 m/s (the negative gradient shows the cyclist is moving back towards home).

(b) Total distance = 600 m and total time = 150 s, so average speed = 600 ÷ 150 = 4 m/s.

(c) The cyclist ends where they started, so the displacement is zero and the average velocity is 0 m/s. This is a favourite exam trap: average speed and average velocity are not the same thing.

Velocity–Time Graphs

A velocity–time graph plots velocity on the vertical axis against time. It carries more information than a distance–time graph, because it gives both the acceleration and the distance.

The gradient gives the acceleration

Gradient = change in velocity ÷ time = acceleration.

A horizontal line means constant velocity: zero acceleration. (A horizontal line on the time axis itself means the object is at rest.)
A straight line sloping upwards means uniform acceleration.
A straight line sloping downwards means uniform deceleration.
A curve means the acceleration is changing. The gradient of the tangent gives the acceleration at that instant.

The area under the graph gives the distance

The area under the graph between two times equals the distance travelled (strictly, the displacement) in that interval. Split the area into rectangles, triangles and trapeziums:

Rectangle: area = base × height.
Triangle: area = ½ × base × height.
Trapezium: area = ½ × (sum of parallel sides) × width.

The trapezium rule is just the equation s = ½(u + v)t in graphical form, which is why motion graphs and the equations of motion always agree.

Worked example 2

A car starts from rest and accelerates uniformly to 20 m/s in 10 s. It travels at this velocity for 30 s, then decelerates uniformly to rest in 5 s. Sketch the velocity–time graph and find (a) the acceleration, (b) the deceleration, (c) the total distance and (d) the average speed.

The graph is a trapezium: a line from (0, 0) to (10, 20), a horizontal line to (40, 20), and a line down to (45, 0).

(a) Acceleration = 20 ÷ 10 = 2 m/s².

(b) Deceleration = 20 ÷ 5 = 4 m/s².

(c) Distance = area of the trapezium. Splitting it: ½ × 10 × 20 = 100 m, plus 30 × 20 = 600 m, plus ½ × 5 × 20 = 50 m. Total = 750 m.

Using the trapezium formula directly: ½ × (30 + 45) × 20 = 750 m, where 30 s and 45 s are the two parallel sides.

(d) Average speed = 750 ÷ 45 ≈ 16.7 m/s.

Worked example 3

A body's velocity increases uniformly from 5 m/s to 15 m/s in 4 s. Find its acceleration and the distance covered.

Acceleration = (15 − 5) ÷ 4 = 2.5 m/s².

The area under the graph is a trapezium with parallel sides 5 and 15 and width 4: ½ × (5 + 15) × 4 = 40 m.

Check with an equation of motion: s = ut + ½at² = 5 × 4 + ½ × 2.5 × 16 = 20 + 20 = 40 m.

Motion Under Gravity

A ball thrown vertically upwards at 20 m/s slows down at g = 10 m/s², stops at the top, then falls back. Taking upwards as positive, its velocity–time graph is a single straight line of gradient −10 m/s², from +20 m/s at t = 0 to −20 m/s at t = 4 s.

The line crosses the time axis at t = 20 ÷ 10 = 2 s. This is the moment the ball is at its highest point.
The maximum height is the area of the triangle above the axis: ½ × 2 × 20 = 20 m.
The area below the axis (from 2 s to 4 s) is also 20 m, but it counts as negative displacement. The total displacement after 4 s is zero: the ball is back in the thrower's hand.

Notice that the gradient stays −10 m/s² even at the top. The velocity is zero for an instant, but the acceleration is not.

Acceleration–Time Graphs

An acceleration–time graph is less common, but the rule is similar: the area under an acceleration–time graph equals the change in velocity.

For example, if a body starting from rest has a constant acceleration of 3 m/s² for 4 s, the graph is a horizontal line at 3 m/s², and the area 3 × 4 = 12 gives a final velocity of 12 m/s.

Summary of Motion Graphs

Distance–time graph: gradient = speed. Horizontal line = at rest. Straight line = uniform speed. Curve = changing speed.
Velocity–time graph: gradient = acceleration, area = distance. Horizontal line = uniform velocity. Straight sloping line = uniform acceleration.
Acceleration–time graph: area = change in velocity.

Exam Tips and Common Mistakes

Read the axes first. Many students lose marks by treating a velocity–time graph as a distance–time graph. A horizontal line means "at rest" on one and "steady velocity" on the other.
Check the units on each axis. If time is in minutes or velocity in km/h, convert before calculating.
Use the whole area. When a graph has several stages, find each area separately and add them.
Average speed is total distance ÷ total time, including any time spent at rest. It is not the average of the speeds.
Use large triangles when finding a gradient from a drawn graph, to reduce reading errors.

Motion Graphs: Practice Questions

A car moves at a constant 25 m/s for 12 s. How far does it travel?
A distance–time graph is a straight line through (0 s, 0 m) and (20 s, 160 m). Find the speed.
A body accelerates uniformly from rest to 30 m/s in 6 s. Find its acceleration and the distance covered.
A lorry decelerates uniformly from 24 m/s to rest in 8 s. Find the deceleration and the stopping distance.
A train accelerates from rest to 12 m/s in 4 s, travels at 12 m/s for 10 s, then comes to rest in 6 s. Find the total distance and the average speed.

Answers: 1. 25 × 12 = 300 m. 2. 160 ÷ 20 = 8 m/s. 3. a = 5 m/s², distance = ½ × 6 × 30 = 90 m. 4. Deceleration = 3 m/s², distance = ½ × 8 × 24 = 96 m. 5. Distance = 24 + 120 + 36 = 180 m, average speed = 180 ÷ 20 = 9 m/s.

Conclusion

Motion graphs turn calculations into geometry. On a distance–time graph, the gradient is the speed. On a velocity–time graph, the gradient is the acceleration and the area under the graph is the distance travelled. Always read the axes and units before you start, split complicated areas into simple shapes, and remember that average speed uses the total distance and the total time.