Sketching Travel Graphs (Cambridge (CIE) O Level Additional Maths): Revision Note
Sketching Travel Graphs
How are s-t, v-t, and a-t graphs related?
Recall that:
Velocity, v, is the rate of change of displacement, s, with respect to time
Acceleration, a, is the rate of change of velocity, v, with respect to time
Differentiate to go from s to v and from v to a
Integrate to go from a to v and from v to s
There will be a constant of integration, c, each time you integrate
![differentiate to go from displacement, to velocity, to acceleration. Integrate to go in the opposite direction.](https://cdn.savemyexams.com/cdn-cgi/image/f=auto,width=3840/https://cdn.savemyexams.com/uploads/2022/01/2.2.1-Using-Calculus-in-1D-Diagram-1.png)
On a velocity-time graph:
Acceleration is the gradient which is found using differentiation
Displacement is the area under the graph which is found using integration
This can also be seen from the units:
![vt graph with tangent to show gradient=acceleration, and shaded area to show integral is the displacement](https://cdn.savemyexams.com/cdn-cgi/image/f=auto,width=3840/https://cdn.savemyexams.com/uploads/2022/03/2-2-1-using-calculus-in-1d-diagram-2.png)
On a displacement-time graph:
Velocity is the gradient which is found using differentiation
The area has no significant meaning
This can also be seen from the units:
On an acceleration-time graph:
Velocity is the area under the graph which is found using integration
The gradient is generally not used
It is a measure called 'jolt' but this is beyond the scope of this course
This can also be seen from the units:
How can I use one travel graph to draw another?
Using the relations stated above, we can inspect either the graph or the equation of the graph, in order to sketch a related travel graph
For example, if a velocity-time graph is a series of sections, with mostly straight lines:
Find the gradient of each section to plot the acceleration-time graph
Remember if the gradient is negative, the acceleration-time graph will be below the x-axis
Find the area underneath each section to help plot the displacement-time graph
Remember that if the velocity is a positive constant (a horizontal line above the x-axis), the displacement will be increasing (a line with positive gradient)
If the velocity is a negative constant (a horizontal line below the x-axis), the displacement will be decreasing (a line with negative gradient)
If a graph is a curve with a known equation, we can use calculus to find the equations of the other related functions
Remember you may need extra information about the velocity or displacement at a point in time when integrating
Once the equations of the other functions are found, they can be sketched
For example if the graph of the displacement-time graph is a cubic
the velocity-time graph will be a quadratic graph
and the acceleration-time graph will be a linear graph
Corresponding s-t, v-t, and a-t graphs for the same journey
![a displacement-time, velocity-time, and acceleration-time graph each showing the same journey](https://cdn.savemyexams.com/cdn-cgi/image/f=auto,width=3840/https://cdn.savemyexams.com/uploads/2023/09/distance-velocity-acceleration.png)
Examiner Tips and Tricks
Questions may involve both differentiation and integration (or finding gradients and areas)
take a moment to double check you have selected the correct method!
Worked Example
A particle moves in a straight line. Its displacement, metres, from a fixed point at time,
seconds, is given by
for
.
Sketch its displacement-time, velocity-time, and acceleration-time graphs.
To sketch the displacement; it can be factorised
The roots can then be found
when:
(repeated root)
and
The graph can then be sketched, noting that it is a negative cubic, and remembering the restriction on the domain;
![displacement time graph for the worked example](https://cdn.savemyexams.com/cdn-cgi/image/f=auto,width=3840/https://cdn.savemyexams.com/uploads/2023/08/kinematics-st-graph.png)
To find the velocity, differentiate the displacement with respect to (time)
This can be factorised to
The roots can then be found
when:
and
The graph can then be sketched, noting that it is a negative quadratic, and remembering the restriction on the domain;
![velocity time graph for the worked example](https://cdn.savemyexams.com/cdn-cgi/image/f=auto,width=3840/https://cdn.savemyexams.com/uploads/2023/08/kinematics-vt-graph.png)
To find the acceleration, differentiate the velocity with respect to This is also the second derivative of the displacement
The graph can then be sketched This is a straight line with -intercept 24, and gradient -12 Remember the restriction on the domain;
![acceleration time graph for the worked example](https://cdn.savemyexams.com/cdn-cgi/image/f=auto,width=3840/https://cdn.savemyexams.com/uploads/2023/08/kinematics-at-graph.png)
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