Velocity-Time Graphs Questions, Worksheets and Revision

Velocity-Time Graphs Questions, Worksheets and Revision

GCSE 6 - 7AQAEdexcelOCRWJECHigherAQA 2022Edexcel 2022OCR 2022WJEC 2022

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:

Level 6-7 GCSE AQA Edexcel OCR WJEC

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

velocity time graph sectionsLevel 6-7GCSEAQAEdexcelOCRWJEC
Level 6-7 GCSE AQA Edexcel OCR WJEC

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.

velocity time graph calculating distance acceleration

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.

Level 6-7GCSEAQAEdexcelOCRWJEC

Skill 2: Calculating Acceleration

Acceleration is calculated as the change in speed over time.

velocity time graph calculating distance acceleration
velocity time graph calculating distance acceleration

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}.

Level 6-7GCSEAQAEdexcelOCRWJEC

Skill 3: Calculating total distance travelled

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

velocity time graph calculating distance acceleration
velocity time graph calculating distance acceleration

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

Level 8-9GCSEAQAEdexcelOCRWJEC
Level 8-9 GCSE AQA Edexcel OCR WJEC

Skill 4: Average of curved graphs

finding the average gradient of a distance time graph
finding the average gradient of a distance time graph

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

finding the average gradient of a distance time graphLevel 6-7GCSEAQAEdexcelOCRWJEC

Skill 5: Instantaneous gradient of a curve

finding the average gradient of a distance time graph
finding the average gradient of a distance time graph

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).

instantaneous acceleration gradient of a curveLevel 6-7GCSEAQAEdexcelOCRWJEC

Example Questions

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.

 

velocity time graphs example 1 answer

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.

 

velocity time graphs example 2 answer

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

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

Related Topics

MME

Areas of Shapes

Level 4-5GCSEKS3
MME

Gradients of Straight Line Graphs

Level 4-5GCSEKS3
MME

Drawing Straight Line Graphs

Level 1-3GCSEKS3

Worksheet and Example Questions

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