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Velocity-Time Graph

Key Stage 4

Meaning

A velocity-time graph is a graph that shows how the velocity of an object changes with time.

About Velocity Time Graphs

Velocity-time graphs give information about the journey taken by an object.
On a velocity-time graph the velocity is plotted on the y-axis and the time is plotted on the x-axis.
A velocity-time graph can be used to calculate the acceleration of an object or the distance travelled by the object.
The gradient of a velocity-time graph is the same as the acceleration.
The area under the curve on a velocity-time graph is the distance travelled by an object.
Constant Velocity Accelerating Decelerating
VtGraphConstantv.png
VtGraphAccelerating.png
VtGraphDecelerating.png
A gradient of zero shows the object is travelling at a constant velocity. Acceleration is shown by a positive gradient. Deceleration is shown by a negative gradient.

Example Calculations

Calculating Acceleration

The acceleration can be calculated from a velocity-time graph by reading the graph and using the equation \(a=\frac{v-u}{t}\).
Calculate the acceleration of the object in this journey. Calculate the acceleration of the object in this journey.
VtGraphCalculateAcceleration1.png
VtGraphCalculateAcceleration2.png
State the known variables.

v = 40m/s

u = 20m/s

t = 8s

State the known variables.

v = 10m/s

u = 40m/s

t = 8s

2. Substitute the numbers into the equation and solve.

\(a = \frac{v-u}{t}\)

\(a = \frac{40-20}{8}\)

\(a = \frac{20}{8}\)

\(a = 2.5m/s/s\)

2. Substitute the numbers into the equation and solve.

\(a = \frac{v-u}{t}\)

\(a = \frac{10-40}{8}\)

\(a = \frac{-30}{8}\)

\(a = -3.75m/s/s\)

Calculate the acceleration of the object at each stage in this journey.
VtGraph1.png
A B C
State the known variables.
VtGraphCalculateAcceleration3.png

v = 30m/s

u = 0m/s

t = 2s

State the known variables.
VtGraphCalculateAcceleration3.png

v = 30m/s

u = 30m/s

t = 3s

State the known variables.
VtGraphCalculateAcceleration3.png

v = 50m/s

u = 30m/s

t = 3s

2. Substitute the numbers into the equation and solve.

\(a = \frac{v-u}{t}\)

\(a = \frac{30-0}{2}\)

\(a = \frac{30}{2}\)

\(a = 15m/s/s\)

2. Substitute the numbers into the equation and solve.

\(a = \frac{v-u}{t}\)

\(a = \frac{30-30}{3}\)

\(a = \frac{0}{3}\)

\(a = 0m/s/s\)

2. Substitute the numbers into the equation and solve.

\(a = \frac{v-u}{t}\)

\(a = \frac{50-30}{3}\)

\(a = \frac{20}{3}\)

\(a = 6.7m/s/s\)

Calculating Distance Travelled

The distance travelled can be calculated from a velocity-time graph by breaking the graph into simple shapes and finding the area of those shapes. This may use the equations \(area = base \times height\) for rectangular shapes and \(area = \frac{base \times height}{2}\) for triangular shapes.
Calculate the distance travelled by the object in this journey. Calculate the distance travelled by the object in this journey.
VtGraphCalculateArea1.png
VtGraphCalculateArea2.png
1. State the known quantities

base = 8s

height = 30m/s

1. State the known quantities

base = 8s

height = 40m/s

2. Substitute the numbers into the equation and solve.

\(area = b \times h\)

\(area = 8 \times 30\)

\(area = distance = 250m\)

2. Substitute the numbers into the equation and solve.

\(area = \frac{b \times h}{2}\)

\(area = \frac{8 \times 40}{2}\)

\(area = \frac{320}{2}\)

\(area = distance = 160m\)

Calculate the distance travelled by the object at each stage in this journey.
VtGraph1.png
A B C
State the known variables.
VtGraphCalculateArea3.png

b = 2s

h = 30m/s

State the known variables.
VtGraphCalculateArea3.png

b = 3s

h = 30m/s

State the known variables.
VtGraphCalculateArea3.png

b = 3s

h1 = 30m/s

h2 = 50m/s

2. Substitute the numbers into the equation and solve.

\(area = \frac{b \times h}{2}\)

\(area = \frac{2 \times 30}{2}\)

\(area = \frac{60}{2}\)

\(area = distance = 30m\)

2. Substitute the numbers into the equation and solve.

\(area = b \times h\)

\(area = 3 \times 30\)

\(area = distance = 90m\)

2. Substitute the numbers into the equation and solve.

\(area = b \times h\)

\(area = 3 \times 30\)

\(area = distance = 90m\) for yellow shaded area.

\(area = \frac{b \times h}{2}\)

\(area = \frac{3 \times (50-30)}{2}\)

\(area = \frac{60}{2}\)

\(area = distance = 30m\) for red area.

Total distance = 120m

References

AQA

Velocity-time graph, pages 148-9, GCSE Physics; Student Book, Collins, AQA
Velocity-time graphs, page 211, GCSE Combined Science; The Revision Guide, CGP, AQA
Velocity-time graphs, pages 136, 138-141, GCSE Physics; Third Edition, Oxford University Press, AQA
Velocity-time graphs, pages 152-3, GCSE Physics, Hodder, AQA
Velocity-time graphs, pages 155-157, GCSE Combined Science Trilogy; Physics, CGP, AQA
Velocity-time graphs, pages 186-188, GCSE Physics; The Complete 9-1 Course for AQA, CGP, AQA
Velocity-time graphs, pages 230-1, 234, GCSE Combined Science Trilogy 2, Hodder, AQA
Velocity-time graphs, pages 62, 63, GCSE Physics; The Revision Guide, CGP, AQA
Velocity-time graphs; area under, pages 156, 157, GCSE Combined Science Trilogy; Physics, CGP, AQA
Velocity-time graphs; area under, pages 187, 188, GCSE Physics; The Complete 9-1 Course for AQA, CGP, AQA
Velocity-time graphs; for falling objects, page 192, GCSE Physics; The Complete 9-1 Course for AQA, CGP, AQA

OCR

Velocity-time graphs, page 162, Gateway GCSE Combined Science; The Revision Guide, CGP, OCR
Velocity-time graphs, page 25, Gateway GCSE Physics; The Revision Guide, CGP, OCR