Uni-Sine
MathematicsMechanics

Velocity-time graphs

A velocity-time graph shows how an object's velocity changes over time, with time on the horizontal axis and velocity on the vertical axis. Two facts unlock every question: the gradient of the line is the acceleration, and the area under the line is the distance travelled (strictly, the displacement). A flat line means constant velocity, and a sloping line means the object is speeding up or slowing down.

Reading a velocity-time graph

Here is a car that pulls away from traffic lights, cruises, then brakes to a stop:

Velocity of the car against time
510152025303540510152020 m/s after 10 sbrakes at 30 sstops at 40 s
Velocity (m/s) against time (s)Drag to pan, scroll to zoom. x from -2 to 42, y from -4 to 24.

Here x is time in seconds and y is velocity in m/s. The graph has three stages:

StageShape of the lineWhat the car is doing
0 to 10 sStraight, sloping upSpeeding up at a constant rate
10 to 30 sFlat (horizontal)Moving at a constant 20 m/s
30 to 40 sStraight, sloping downSlowing down at a constant rate
ShapeMeaning
Horizontal lineConstant velocity, zero acceleration
Horizontal line along the time axisStationary
Straight line sloping upConstant acceleration
Straight line sloping downConstant deceleration
Steeper lineGreater acceleration
Curve getting steeperAcceleration increasing
Curve getting flatterAcceleration decreasing
Line below the time axisMoving in the opposite direction

Gradient of a velocity-time graph: acceleration

Rise over run on this graph means velocity gained per second, which is exactly what acceleration measures:

a = \frac{\Delta v}{t} = \frac{\text{change in velocity}}{\text{time taken}}

Example 1: acceleration of each stage

  • 0 to 10 s: a = (20 − 0) ÷ 10 = 2 m/s².
  • 10 to 30 s: the velocity does not change, so a = 0 m/s².
  • 30 to 40 s: a = (0 − 20) ÷ 10 = −2 m/s². The minus sign means it is slowing down: a deceleration of 2 m/s².

Try reading one yourself in the Laws of Motion course, then press Mark.

Read an acceleration (slope) or a distance (area) off a random velocity-time graph.

Press Mark to have your answer marked in the Laws of Motion course.

Area under a velocity-time graph: distance

Velocity × time = distance, and on the graph velocity × time is an area. So the area under the line, down to the time axis, is the distance travelled. Split the shape into rectangles and triangles, or use the trapezium (trapezoid) area formula.

Example 2: total distance for the car

  • 0 to 10 s, triangle: ½ × 10 × 20 = 100 m.
  • 10 to 30 s, rectangle: 20 × 20 = 400 m.
  • 30 to 40 s, triangle: ½ × 10 × 20 = 100 m.

Total distance = 100 + 400 + 100 = 600 m. You could also treat the whole shape as one trapezium with parallel sides 40 s and 20 s and height 20 m/s: ½ × (40 + 20) × 20 = 600 m.

Example 3: average speed

\text{average speed} = \frac{\text{total distance}}{\text{total time}} = \frac{600}{40} = 15\ \text{m/s}

The SUVAT equations are these same gradient and area facts written as formulas, for any stage with constant acceleration.

Curved velocity-time graphs

If the line is curved, the acceleration is changing. You can still find the gradient and the area, but you have to estimate them.

y = 0.5{ x}^{2}
0.511.522.533.544.555.566.55101520252 m/s8 m/s18 m/s
Drag to pan, scroll to zoom. x from 0 to 6.5, y from -4.23 to 25.4.

Here x is time in seconds and y is velocity in m/s. The curve gets steeper, so the object's acceleration is increasing.

Example 4: estimating the area with strips

Split the area from 0 to 6 s into three strips, each 2 s wide, and treat each as a trapezium. The velocities at 0, 2, 4 and 6 s are 0, 2, 8 and 18 m/s:

\frac{1}{2} \times 2 \times (0 + 2) + \frac{1}{2} \times 2 \times (2 + 8) + \frac{1}{2} \times 2 \times (8 + 18) = 2 + 10 + 26 = 38\ \text{m}

The estimate is 38 m. It is a slight overestimate, because the straight tops of the trapezia sit just above the curve. The exact answer, found with integration at A-level, is 36 m. More, narrower strips give a better estimate. On squared paper, counting squares works too: find the distance one square stands for, then count the squares under the line.

Example 5: estimating acceleration with a tangent

To find the acceleration at one moment on a curve, draw a tangent: a straight line that just touches the curve at that point. Its gradient is the acceleration at that instant. For this curve, the tangent at t = 4 s goes through (4, 8) and (6, 16), so the acceleration is (16 − 8) ÷ (6 − 4) = 4 m/s².

Check your understanding.

practice
Pick an answer, then check it.
On a velocity-time graph, a cyclist's line goes in a straight line from 4 m/s at 0 s to 10 m/s at 3 s. What is the acceleration?

Velocity or speed?

Velocity has a direction, so it can be negative: a line below the time axis means the object is moving backwards. Area below the axis counts as negative displacement. If a question asks for total distance, add the sizes of the areas above and below the axis. If it asks for displacement, subtract the area below from the area above. A speed-time graph never goes below the axis, because speed has no direction.

Common mistakes

  • Mixing up velocity-time and distance-time graphs. The same shape means different things on each.
  • Reading the gradient as distance or the area as acceleration. Gradient → acceleration, area → distance.
  • Forgetting the ½ for triangle areas.
  • Unit slips. If time is in minutes or velocity in km/h, convert before working out the area, or the distance will be in strange units.
  • Dropping the minus sign on a deceleration. A downward slope gives a negative gradient.
  • Calling a flat line "stopped". On a velocity-time graph it means constant velocity.

Practise with a short timed paper

Motion graphs come up in GCSE Physics and GCSE Maths Higher. The daily GCSE Physics Higher and GCSE Maths Higher papers are short timed papers of 8 questions, with a new one every day.

Common questions

What does the gradient of a velocity-time graph show?
The acceleration. Work it out as change in velocity ÷ time taken.
What does the area under a velocity-time graph show?
The distance travelled, or strictly the displacement.
What does a horizontal line on a velocity-time graph mean?
The object is moving at a constant velocity, so its acceleration is zero.
How do you find distance from a curved velocity-time graph?
Estimate the area by splitting it into strips and treating each as a trapezium, or by counting squares.
What does a negative gradient mean?
The object is slowing down, if its velocity is positive: a deceleration.
What is the difference between a velocity-time and a speed-time graph?
Velocity includes direction, so a velocity-time graph can go below the axis. A speed-time graph cannot.

Revision cards

Say the answer first, then flip the card to check it.

Where to go next

  • Solve it

    SUVAT Calculator

    Solve any equation of motion from three known values

    Open the tool
  • Learn it

    Laws of Motion

    Newton's laws, momentum, and the forces that shape our world

    Start the course
  • Test it

    GCSE Physics Higher

    A new exam-style paper every day, with instant marking and a mark scheme

    Start today's paper
Read next
Maths · MechanicsDisplacement-time graph: gradient, velocity and examplesHow to read a displacement-time graph: gradient as velocity, flat, steep and curved lines, tangents, and how it differs from a distance-time graph.