# Velocity-Time Graphs

## Velocity-Time Graphs Revision

**Velocity-Time Graphs**

A **velocity-time graph ** (or speed-time graph) is a way of visually **expressing a journey**.

We are going to be using velocity-time graphs to find two things, primarily: **total distance, **and **acceleration**.

There are **5** key skills you need to learn

Make sure you are happy with the following topics before continuing:

**Velocity-Time Graphs – Key things to remember:**

With speed on the y-axis and time on the x-axis, a speed-time graph tells us how someone/something’s speed has changed over a period of time.

**1) The gradient of the line = Acceleration **

**2) Negative gradient = Deceleration**

**3) Flat section means constant velocity (NOT STOPPED)**

**4) Area under the graph = Distance travelled**

**Skill 1: Describing a graph**

One Skill you will need learn is describing a velocity time graph.

**Example:** The speed-time graph shows a 50-second car journey. Describe the 50 second journey.

**Step 1:** Split the graph up into distinct sections, these can be seen in the image as A, B, C and D.

**Step 2:** In detail describe each part of the journey, ensuring to use numerical values throughout.

Section A – The car **accelerated** from 0 to 15 m/s over the first 10 seconds (because the line is straight, the acceleration is **constant**).

Section B – The line is flat, meaning the car’s speed did not change for 10 seconds – meaning it was moving at a **constant speed**.

Section C – The car **accelerated** up to 25 m/s over the next 10 seconds,

Section D – Finally it spent the last 20 seconds **decelerating **back down to 0 m/s.

**Skill 2: Calculating Acceleration**

Acceleration is calculated as the change in speed over time.

**Example:** The speed-time graph shows a 50-second car journey, find which section of the graph has the greatest acceleration.

We know,

**The gradient of the line = Acceleration**

We must find the gradient of the each section.

**Section** \bf{A}: Acceleration between 0s and 10s = gradient=\dfrac{15-0}{10-0}=1.5 m/s^2

**Section** \bf{B}: This section is flat, meaning the acceleration will be 0

**Section** \bf{C}: Acceleration between 20s and 30s = gradient =\dfrac{25-15}{30-20}=1 m/s^2

**Section** \bf{D}: Acceleration between 30s and 50s = gradient =\dfrac{0-25}{50-30}=\dfrac{-25}{20} = -1.25 m/s^2

Section \bf{A} has the largest acceleration, so the **maximum acceleration** is 1.5 m/s^2

Note: units of acceleration are expressed in **distance/time**\bold{^2}, which in this case is **m/s**\bold{^2}.

**Skill 3: Calculating total distance travelled**

Calculating the total distance travelled is one of the most common exam questions you may see.

**Example:** The speed-time graph shows a 50-second car journey, Calculate the total distance travelled over the 50 seconds.

we know,

**Area under the graph = Distance travelled**

To work out the area under this graph, we will break it into 4 shapes: A, B, C, and D.

This gives two triangles, a rectangle, and a trapezium, which are all shapes that we can work out the area of.

\text{A}=\dfrac{1}{2}\times10 \times15= 75 m

\text{B}=10\times15=150 m

C=\dfrac{1}{2}(15+25)\times10=200 m

D=\dfrac{1}{2}\times20\times25=250 m

**Total distance travelled**:

75+150+200+250=675 m

**Skill 4: Average of curved graphs**

Finding the average gradient, is the gradient over a length of time.

**Example:** A speed-time graph of the first 4 seconds of someone running a race is shown.

Calculate the average acceleration over the 4 seconds.

We know:

**The gradient of the line = Acceleration**

To work out the **average acceleration **over the 4 seconds, we will draw a line from where the graph is at 0 s to where the graph is at 4 s and find the gradient of it.

So, we get the average acceleration to be,

\text{gradient}=\dfrac{6-0}{4-0}=1.5 m/s^2

**Skill 5: Instantaneous gradient of a curve**

Finding the instantaneous gradient, is the gradient of the tangent at a point.

**Example:** A speed-time graph of the first 4 seconds of someone running a race is shown.

Calculate the instantaneous acceleration 2 seconds in.

To do this we will draw a **tangent **to the line after 2 seconds and work out the gradient of that. This is shown above.

Then, we get the instantaneous acceleration to be,

\text{gradient}=\dfrac{5.8-3.2}{3.5-1.0} = 1.04 m/s^2 (3 sf).

## Velocity-Time Graphs Example Questions

**Question 1:** A ball is placed at rest at the top of a hill. It travels with constant acceleration for the first 12 second and reaches a speed of 4 m/s. It then **decelerates** at a constant rate of 0.1 m/s^2 for 20 seconds. It then travels at a constant speed for a further 18 seconds.

**[4 marks]**

Draw a speed-time graph for the ball over the course of this 50 seconds.

So, we will firstly draw a straight from the origin to (12, 4), since after 12 seconds, it’s reached 4 m/s. Then, for the next part we’re told the deceleration is 0.1 m/s^2 for 20 seconds. So, if the speed decreases by 0.1 every second, after 20 seconds it will be

0.1 \times 20=2 m/s

Therefore, by 32 seconds in the speed is 2 m/s, so we will draw a straight line from (12, 4) to (32, 2). Finally, a constant speed will be represented by a flat line that goes until the 50 second point, still at 2 m/s. The result should look like the graph below.

**Question 2:** Below is a speed-time graph of a track cyclist during a race. Work out the total distance travelled by the cyclist over the course of the race.

**[3 marks]**

We need to find the area underneath the graph. To do this, we will split it up into shapes we know how to calculate the area of, as seen below.

A is a triangle, B and C are trapeziums, and D is a rectangle. So, we get

\text{A}=\dfrac{1}{2}\times10\times15=75 m

\text{B}=\dfrac{1}{2}\times(10+15)\times5=62.5 m

\text{C}=\dfrac{1}{2}\times(10+20)\times5=75 m

\text{D}=30\times20 = 600 m

Therefore, the total distance travelled by the cyclist is

75+62.5+75+600=812.5 m

**Question 3:** Below is a speed-time graph of a runner during the first 50 seconds of a race. Work out the average acceleration of the runner during this period.

**[2 marks]**

In order to determine the average acceleration, we draw a line from the origin to the endpoint of the graph, as seen below. The average acceleration is given by the gradient of this line.

Hence the average acceleration is,

\text{gradient}=\dfrac{4-0}{50-0}= 0.08 m/s^2