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

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.

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.
colspan = "3"
State the known variables. v = 40m/s u = 20m/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. \(a = \frac{v-u}{t}\) \(a = \frac{10-40}{8}\) \(a = \frac{-30}{8}\) \(a = -3.75m/s/s\)

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.

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. \(area = \frac{b \times h}{2}\) \(area = \frac{8 \times 40}{2}\) \(area = \frac{320}{2}\) \(area = distance = 160m\)