## What you need to know

## 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**.

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 steady velocity (NOT STOPPED)**

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

## Example 1 – Acceleration and Distance travelled

The speed-time graph below describes a 50-second car journey.

Work out the total distance travelled by the car and work out the maximum acceleration of the car during the 50 seconds.

**[4 marks]**

Lets analyse the graph:

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

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

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

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

**Work out the total distance travelled by the car:**

**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 -two triangles, a rectangle, and a trapezium are all shapes that we can work out the area of. So, we get

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

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

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

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

**Total distance travelled**:

75+150+200+250=675\text{m}

**Work out the maximum acceleration:**

**The gradient of the line = Acceleration**

First lest find the gradient of the each section.

**Section** \bf{A} :

\text{acceleration between 0s and 10s = gradient}=\dfrac{15-0}{10-0}=1.5\text{m/s}^2

**Section** \bf{B}

This section is flat, meaning meaning the acceleration will be 0

**Section** \bf{C}

\text{acceleration between 20s and 30s = gradient}=\dfrac{25-15}{30-20}=1\text{m/s}^2

**Section** \bf{D}

\text{acceleration between 30s and 50s = gradient}=\dfrac{0-25}{50-30}=\dfrac{-25}{20} = -1.25\text{m/s}^2

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

**Example 2**

Below is a speed-time graph of the first 4 seconds of someone running a race.

Work out the average acceleration over the 4 seconds and work out the instantaneous acceleration 2 seconds in.

we know:

**The gradient of the line = Acceleration**

### Work out the average acceleration over the 4 seconds

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

So, we get that average acceleration to be,

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

## work out the instantaneous acceleration 2 seconds in

To so 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\text{ m/s}^2\text{ (3 s.f.)}

### 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\text{ m/s}^2 for 20 seconds. It then travels at a constant speed for a further 18 seconds.

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\text{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\text{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.

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\text{m}

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

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

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

Therefore, the total distance travelled by the cyclist is

75+62.5+75+600=812.5\text{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.

In order to determine the average acceleration we draw a line from the origin to the endpoint of the graph. The average acceleration sis given by the gradient of this line.

Hence the average acceleration is,

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

### Worksheets and Exam Questions

#### (NEW) Velocity-Time Graphs Exam Style Questions - MME

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