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Calculus for Kinematics (Cambridge (CIE) IGCSE Additional Maths): Revision Note

Exam code: 606

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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

Worked Example

The displacement, x m, of a particle at t seconds, is modelled by the function x left parenthesis t right parenthesis equals 2 t cubed minus 27 t squared plus 84 t.

Find expressions for x apostrophe open parentheses t close parentheses and x apostrophe apostrophe open parentheses t close parentheses, and state what these expressions represent.

x equals 2 t cubed minus 27 t squared plus 84 t

x apostrophe open parentheses t close parentheses means the same as fraction numerator d x over denominator d t end fraction

x apostrophe open parentheses t close parentheses space equals space 6 t squared minus 54 t plus 84 x apostrophe open parentheses t close parentheses space equals space 6 open parentheses t squared minus 9 t plus 14 close parentheses


This expression represents the velocity of the particle.
Answers may or may not need to be factorised, depending on the question

x apostrophe apostrophe open parentheses t close parentheses means the same as fraction numerator d squared x over denominator d t squared end fraction

x apostrophe apostrophe open parentheses t close parentheses space equals space 12 t minus 54


This expression represents the acceleration of the particle.

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

      • You could think of this as the "net area" e.g. 4 m - 1 m = 3 m

    • 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

      • You could think of this as the "gross area" e.g. 4 m + 1 m = 5 m

Examiner Tips and Tricks

  • 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 ms-1 at time t seconds modelled by space v left parenthesis t right parenthesis equals 8 t cubed minus 12 t squared minus 2 t.

a) Given that the initial position of the particle is at the origin, find an expression for its displacement from the origin at time t seconds.

 

The integral of the velocity gives the displacement

 

s open parentheses t close parentheses space equals integral v open parentheses t close parentheses space d t space space equals space integral open parentheses 8 t cubed minus 12 t squared minus 2 t close parentheses space d t s open parentheses t close parentheses space equals space 2 t to the power of 4 minus 4 t cubed minus t squared plus c

 

"Initial" means when t equals 0 and "at the origin" means s equals 0

Substitute these values in to find c

 

0 space equals space 2 open parentheses 0 close parentheses to the power of 4 minus 4 open parentheses 0 close parentheses cubed minus open parentheses 0 close parentheses squared plus c c equals 0

 

bold italic s open parentheses t close parentheses space equals space 2 bold italic t to the power of 4 minus 4 bold italic t cubed minus bold italic t squared

 

b) Find the displacement of the particle from the origin after the first five seconds of its motion.

 

To find the displacement after 5 seconds, integrate the velocity between 0 and 5

 

integral subscript 0 superscript 5 8 t cubed minus 12 t squared minus 2 t space d t open square brackets 2 t to the power of 4 minus 4 t cubed minus t squared close square brackets subscript 0 superscript 5 space equals space open square brackets 2 open parentheses 5 close parentheses to the power of 4 minus 4 open parentheses 5 close parentheses cubed minus open parentheses 5 close parentheses squared close square brackets minus open square brackets 2 open parentheses 0 close parentheses to the power of 4 minus 4 open parentheses 0 close parentheses cubed minus open parentheses 0 close parentheses squared close square brackets equals 725 minus 0

 

bold 725 bold space bold m

 

c) Explain why this value is not the same as the distance travelled during the first 5 seconds.

 

When finding the displacement, any negative displacements are taken into account. e.g. 3m - 1m = 2m. When finding the distance, we are looking for the total of all the areas, ignoring their direction/sign. e.g. 3m + 1m = 4m. This function for velocity has some regions underneath the x-axis, and some regions above the x-axis between 0 and 5 seconds, so the displacement and distance covered in this time will be different.

 

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