Velocity-Time Graphs

A speed-time graph shows how the speed of an object varies with time
A velocity-time graph shows how the speed of an object moving in a certain direction varies with time

Acceleration & Deceleration on a Speed-Time Graph

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The red line shows the speed of an object increasing over time, whilst the green line shows the speed of an object decreasing over time

Acceleration on a Velocity-Time Graph

The slope of the line represents the magnitude of acceleration
- A steep slope means large acceleration (or deceleration)
  - The object's velocity changes very quickly
- A gentle slope means small acceleration (or deceleration)
  - The object's velocity changes very gradually
- A flat line means the acceleration is zero
  - The object is moving with a constant velocity
  - A constant velocity means a constant speed in a straight line

Interpreting Gradients on a Velocity-Time Graphs

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This image shows how to interpret the slope of a velocity-time graph

The acceleration of an object can be calculated from the gradient of a velocity-time graph

$acceleration = gradient = \frac{rise}{run}$

Calculating Gradient of a Velocity-Time Graph

T6_Elhx1_2-1-gradient-velocity-time-graph

How to find the gradient of a velocity-time graph

Worked example

Tora is training for a cycling tournament.

The velocity-time graph below shows her motion as she cycles along a flat, straight road.

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(a) In which section of the speed-time graph is Tora’s acceleration the largest?

(b) Calculate Tora’s acceleration between 5 and 10 seconds.

Answer:

Part (a)

Step 1: Recall that the slope of a speed-time graph represents the magnitude of acceleration

The slope of a speed-time graph indicates the magnitude of acceleration
Therefore, the only sections of the graph where Tora is accelerating is section B and section D
Sections A, C, and E are flat – in other words, Tora is moving at a constant speed (i.e. not accelerating)

Step 2: Identify the section with the steepest slope

Section D of the graph has the steepest slope
Hence, the largest acceleration is shown in section D

Part (b)

Step 1: Recall that the gradient of a speed-time graph gives the acceleration

Calculating the gradient of a slope on a speed-time graph gives the acceleration for that time period

Step 2: Draw a large gradient triangle at the appropriate section of the graph

A gradient triangle is drawn for the time period between 5 and 10 seconds below:

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Step 3: Calculate the size of the gradient and state this as the acceleration

The acceleration is given by the gradient, which can be calculated using:

acceleration = gradient = 5 ÷ 5 = 1 m/s²

Therefore, Tora accelerated at 1 m/s² between 5 and 10 seconds

Exam Tip

Use the entire slope, where possible, to calculate the gradient. Examiners tend to award credit if they see a large gradient triangle used - so remember to draw 'rise' and 'run' lines directly on the graph itself!

Area Under a Velocity-Time Graph

Higher Tier

The distance travelled by an object can be found by determining the area beneath a velocity-time graph

Calculating the Area Under a Velocity-Time Graph

velocity-time-graphs-3, IGCSE & GCSE Chemistry revision notes

The distance travelled can be found from the area beneath the graph

If the area beneath the graph forms a triangle (the object is accelerating or decelerating) then the area can be determined using the formula:

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

If the area beneath the graph is a rectangle (constant velocity) then the area can be determined using the formula:

$area = base \times height$

Worked example

The velocity-time graph below shows a car journey which lasts for 160 seconds.

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Calculate the total distance travelled by the car on this journey.

Answer:

Step 1: Recall that the area under a velocity-time graph represents the distance travelled

To calculate the total distance travelled, the total area underneath the line must be determined

Step 2: Identify each enclosed area

In this example, there are five enclosed areas under the line
These can be labelled as areas 1, 2, 3, 4 and 5, as shown in the image below:

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Step 3: Calculate the area of each enclosed shape under the line

Area 1 = area of a triangle = ½ × base × height = ½ × 40 × 17.5 = 350 m
Area 2 = area of a rectangle = base × height = 30 × 17.5 = 525 m
Area 3 = area of a triangle = ½ × base × height = ½ × 20 × 7.5 = 75 m
Area 4 = area of a rectangle = base × height = 20 × 17.5 = 350 m
Area 5 = area of a triangle = ½ × base × height = ½ × 70 × 25 = 875 m

Step 4: Calculate the total distance travelled by finding the total area under the line

Add up each of the five areas enclosed:

total distance = 350 + 525 + 75 + 350 + 875

total distance = 2175 m

Velocity-Time Graphs (WJEC GCSE Physics)

Revision Note