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.

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

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

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

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

Exam Tip

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 $s = - 2 t^{3} + 12 t^{2}$ for $0 \leq t \leq 7$ .

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

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

$s = - 2 t^{3} + 12 t^{2} = - 2 t^{2} (t - 6)$

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; $0 \leq t \leq 7$

displacement time graph for the worked example

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

$v = \frac{d s}{d t} = - 6 t^{2} + 24 t$

This can be factorised to

$v = - 6 t (t - 4)$

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; $0 \leq t \leq 7$

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

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

The graph can then be sketched
This is a straight line with $y$ -intercept 24, and gradient -12
Remember the restriction on the domain; $0 \leq t \leq 7$

acceleration time graph for the worked example

Sketching Travel Graphs (CIE IGCSE Additional Maths)

Revision Note