Sketching Travel Graphs (Cambridge International IGCSE Further Maths)

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

• 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:

  • $\mathrm{Gradient}=\frac{\mathrm{ms}^{-1}}{s}=\mathrm{ms}^{-2}=\mathrm{Acceleration}$
  • $\mathrm{Area}=\mathrm{ms}^{-1}\times s=m=\mathrm{Displacement}$

• 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:

  • $\mathrm{Gradient}=\frac{m}{s}=\mathrm{ms}^{-1}=\mathrm{Velocity}$

• 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:

  • $\mathrm{Area}=\mathrm{ms}^{-2}\times s=\mathrm{ms}^{-1}=\mathrm{Velocity}$

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

Exam Hint

• 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 $s=-2t^{3}+12t^{2}$ for $0\leq t\leq7$.

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

To sketch the displacement; $s=-2t^{3}+12t^{2}$ it can be factorised

s=−2t³+12t²=−2t² (t−6)

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; $0\leq t\leq7$

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

$v=\frac{ds}{dt}=-6t^{2}+24t$

This can be factorised to

v=−6t (t−4)

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; $0\leq t\leq7$

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

$a=\frac{d^{2}s}{dt^{2}}=\frac{dv}{dt}=-12t+24$

The graph can then be sketched This is a straight line with $y$-intercept 24, and gradient -12 Remember the restriction on the domain; $0\leq t\leq7$