IGCSEFurther MathsCambridge (CIE)Additional MathsRevision NotesCalculusCalculus for KinematicsSketching Travel Graphs

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

Exam code: 606

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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 space equals fraction numerator ms to the power of negative 1 end exponent space over denominator straight s end fraction equals ms to the power of negative 2 end exponent equals Acceleration

    • Area space equals ms to the power of negative 1 end exponent cross times straight s space equals straight m equals 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 space equals fraction numerator straight m space over denominator straight s end fraction equals ms to the power of negative 1 end exponent equals 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 space equals ms to the power of negative 2 end exponent cross times straight s space equals ms to the power of negative 1 end exponent equals 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 equals negative 2 t cubed plus 12 t squared for 0 less or equal than t less or equal than 7.

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

To sketch the displacement; s equals negative 2 t cubed plus 12 t squared it can be factorised

s equals negative 2 t cubed plus 12 t squared equals negative 2 t squared open parentheses t minus 6 close parentheses

The roots can then be found

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

The graph can then be sketched, noting that it is a negative cubic, and remembering the restriction on the domain; 0 less or equal than t less or equal than 7

displacement time graph for the worked example

 

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

v equals fraction numerator d s over denominator d t end fraction equals negative 6 t squared plus 24 t

This can be factorised to

v equals negative 6 t open parentheses t minus 4 close parentheses

The roots can then be found

v equals 0 when:
t equals 0 and t equals 4

The graph can then be sketched, noting that it is a negative quadratic, and remembering the restriction on the domain; 0 less or equal than t less or equal than 7

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 equals fraction numerator d squared s over denominator d t squared end fraction equals fraction numerator d v over denominator d t end fraction equals negative 12 t plus 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; 0 less or equal than t less or equal than 7

acceleration time graph for the worked example
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