Sketching Travel Graphs (Cambridge (CIE) O Level Additional Maths): Revision Note

Exam code: 4037

Dan Finlay

Written by: Dan Finlay

Reviewed by: Lucy Kirkham

Updated on

Sketching travel graphs

  • 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.
  • 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:

    • Gradient =ms1 s=ms2=Acceleration

    • Area =ms1×s =m=Displacement

vt graph with tangent to show gradient=acceleration, and shaded area to show integral is the displacement
  • 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:

    • Gradient =m s=ms1=Velocity

  • 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:

    • Area =ms2×s =ms1=Velocity

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

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, s metres, from a fixed point at time, t seconds, is given by s=2t3+12t2 for 0t7.

Sketch its displacement-time, velocity-time, and acceleration-time graphs.

To sketch the displacement; s=2t3+12t2 it can be factorised

s=2t3+12t2=2t2(t6)

The roots can then be found

s=0 when:
t=0 (repeated root)
and t=6

The graph can then be sketched, noting that it is a negative cubic, and remembering the restriction on the domain; 0t7

displacement time graph for the worked example

 

To find the velocity, differentiate the displacement with respect to t (time)

v=dsdt=6t2+24t

This can be factorised to

v=6t(t4)

The roots can then be found

v=0 when:
t=0 and t=4

The graph can then be sketched, noting that it is a negative quadratic, and remembering the restriction on the domain; 0t7

velocity time graph for the worked example

To find the acceleration, differentiate the velocity with respect to t This is also the second derivative of the displacement

a=d2sdt2=dvdt=12t+24

The graph can then be sketched This is a straight line with y-intercept 24, and gradient -12 Remember the restriction on the domain; 0t7

acceleration time graph for the worked example

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Dan Finlay

Author: Dan Finlay

Expertise: Portfolio Lead

Dan graduated from the University of Oxford with a First class degree in mathematics. As well as teaching maths for over 8 years, Dan has marked a range of exams for Edexcel, tutored students and taught A Level Accounting. Dan has a keen interest in statistics and probability and their real-life applications.

Lucy Kirkham

Reviewer: Lucy Kirkham

Expertise: Content Creator

Lucy has been a passionate Maths teacher for over 12 years, teaching maths across the UK and abroad helping to engage, interest and develop confidence in the subject at all levels.Working as a Head of Department and then Director of Maths, Lucy has advised schools and academy trusts in both Scotland and the East Midlands, where her role was to support and coach teachers to improve Maths teaching for all.