DP IB Maths: AA SL

Revision Notes

5.6.2 Calculus for Kinematics

Test Yourself

Differentiation for Kinematics

How is differentiation used in kinematics?

  • Displacement, velocity and acceleration are related by calculus
  • In terms of differentiation and derivatives
    • velocity is the rate of change of displacement
      • v equals fraction numerator straight d s over denominator straight d t end fraction  or  v left parenthesis t right parenthesis equals s apostrophe left parenthesis t right parenthesis
    • acceleration is the rate of change of velocity
      • a equals fraction numerator straight d v over denominator straight d t end fraction  or  a left parenthesis t right parenthesis equals v apostrophe left parenthesis t right parenthesis
    • so acceleration is also the second derivative of displacement
      • a equals fraction numerator straight d squared s over denominator straight d t squared end fraction  or  a left parenthesis t right parenthesis equals s apostrophe apostrophe left parenthesis t right parenthesis
  • If a graph is not given you can use your GDC to draw one
    • you can then use your GDC’s graphing features to find gradients
      • velocity is the gradient on a displacement (-time) graph
      • acceleration is the gradient on a velocity (-time) graph

Worked example

The displacement,space s space straight m, of a particle atspace t seconds, is modelled by s left parenthesis t right parenthesis equals 2 t cubed minus 27 t squared plus 84 t

  1. Findspace v left parenthesis t right parenthesis andspace a left parenthesis t right parenthesis.
  2. Find the times at which the particle is at rest.

5-6-2-ib-sl-aa-only-diffk-we-soltn

Integration for Kinematics

 How is integration used in kinematics?

  • Since velocity is the derivative of displacement (v equals fraction numerator straight d s over denominator straight d t end fraction) it follows that

space s equals integral v space straight d t

  • Similarly, velocity will be an antiderivative of acceleration

space v equals integral a space straight d t

How would I find the constant of integration in kinematics problems?

  • A boundary or initial condition would need to be known
    • phrases involving the word “initial”, or “initially” are referring to time being zero, i.e. space t equals 0
    • you might also be given information about the object at some other time (this is called a boundary condition)
    • substituting the values in from the initial or boundary condition would allow the constant of integration to be found

How are definite integrals used in kinematics?

  • Definite integrals can be used to find the displacement of a particle between two points in time
    • space integral subscript t subscript 1 end subscript superscript t subscript 2 end superscript space v left parenthesis t right parenthesis space straight d t would give the displacement of the particle between the timesspace t equals t subscript 1 andspace t equals t subscript 2
      • This can be found using a velocity-time graph by subtracting the total area below the horizontal axis from the total area above
    • space integral subscript t subscript 1 end subscript superscript t subscript 2 end superscript open vertical bar v left parenthesis t right parenthesis close vertical bar space straight d t gives the distance a particle has travelled between the timesspace t equals t subscript 1 andspace t equals t subscript 2
      • This can be found using a velocity velocity-time graph by adding the total area below the horizontal axis to the total area above
      • Use a GDC to plot the modulus graphspace y equals open vertical bar v left parenthesis t right parenthesis close vertical bar

5-6-2-ib-sl-aa-only-kin-mod-graph

Exam Tip

  • Sketching the velocity-time graph can help you visualise the distances travelled using areas between the graph and the horizontal axis

Worked example

A particle moving in a straight horizontal line has velocity (v space straight m space straight s to the power of negative 1 end exponent) at timespace t seconds modelled byspace v left parenthesis t right parenthesis equals 8 t cubed minus 12 t squared minus 2 t.

  1. Given that the initial position of the particle is at the origin, find an expression for its displacement from the origin at timespace t seconds.
  2. Find the displacement of the particle from the origin in the first five seconds of its motion.
  3. Find the distance travelled by the particle in the first five seconds of its motion.

5-6-2-ib-sl-aa-only-int-kin-we-soltn

You've read 0 of your 0 free revision notes

Get unlimited access

to absolutely everything:

  • Downloadable PDFs
  • Unlimited Revision Notes
  • Topic Questions
  • Past Papers
  • Model Answers
  • Videos (Maths and Science)

Join the 80,663 Students that ❤️ Save My Exams

the (exam) results speak for themselves:

Did this page help you?

Paul

Author: Paul

Paul has taught mathematics for 20 years and has been an examiner for Edexcel for over a decade. GCSE, A level, pure, mechanics, statistics, discrete – if it’s in a Maths exam, Paul will know about it. Paul is a passionate fan of clear and colourful notes with fascinating diagrams – one of the many reasons he is excited to be a member of the SME team.