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Calculus for Kinematics (Edexcel IGCSE Further Pure Maths) : Revision Note

Roger B

Written by: Roger B

Reviewed by: Dan Finlay

Updated on

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  (differentiate displacement to get velocity)

    • acceleration is the rate of change of velocity

      • a equals fraction numerator straight d v over denominator straight d t end fraction  (differentiate velocity to get acceleration)

    • 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  (differentiate displacement twice to get acceleration)

  • On a graph this means that

    • velocity is the gradient on a displacement-time graph

    • acceleration is the gradient on a velocity-time) graph

  • You can also use this to find (local) minimum and maximum values

    • Exactly the same as using calculus to find other minimum and maximum values

    • e.g. to find any local minimum or maximum values for velocity

      • look for times t at which  fraction numerator straight d v over denominator straight d t end fraction equals 0

Worked Example

The displacement, s metres, of a particle at time t seconds, is modelled by  s left parenthesis t right parenthesis equals 2 t cubed minus 18 t squared plus 48 t,  t greater or equal than 0. 

(a) Find expressions in terms of t for the velocity and acceleration of the particle.

Differentiate the displacement to find the velocity

Note that this 'powers of t' derivative works exactly the same way as a 'powers of x' derivative!

table row v equals cell fraction numerator straight d s over denominator straight d t end fraction end cell row blank equals cell 2 open parentheses 3 t to the power of 3 minus 1 end exponent close parentheses minus 18 open parentheses 2 t to the power of 2 minus 1 end exponent close parentheses plus 48 end cell end table


bold italic v stretchy left parenthesis t stretchy right parenthesis bold equals bold 6 bold italic t to the power of bold 2 bold minus bold 36 bold italic t bold plus bold 48


Now differentiate the velocity to find the acceleration.

table attributes columnalign right center left columnspacing 0px end attributes row a equals cell fraction numerator straight d v over denominator straight d t end fraction end cell row blank equals cell 6 open parentheses 2 t to the power of 2 minus 1 end exponent close parentheses minus 36 end cell end table

bold italic a begin bold style stretchy left parenthesis t stretchy right parenthesis end style bold equals bold 12 bold italic t bold minus bold 36

(b) Find the time(s) at which the particle is instantaneously at rest.

'Instantaneously at rest' means that the velocity is zero

So solve v open parentheses t close parentheses equals 0 for t

table attributes columnalign right center left columnspacing 0px end attributes row cell 6 t squared minus 36 t plus 48 end cell equals 0 row cell 6 open parentheses t squared minus 6 t plus 8 close parentheses end cell equals 0 row cell t squared minus 6 t plus 8 end cell equals 0 row cell open parentheses t minus 2 close parentheses open parentheses t minus 4 close parentheses end cell equals 0 end table
t equals 2 space space space or space space space t equals 4


The particle is at rest at bold italic t bold equals bold 2 seconds and at bold italic t bold equals bold 4 seconds

(c) Find the minimum velocity of the particle.

First of all note that this is a different question from 'find the minimum speed of the particle'
The answer to that would be zero, at the times found in part (b)!


The graph of  v open parentheses t close parentheses equals 6 t squared minus 36 t plus 48  is a 'u-shaped' parabola

So we know there is going to be a single minimum point

That minimum point will occur when fraction numerator straight d v over denominator straight d t end fraction equals 0

So start by solving  a equals fraction numerator straight d v over denominator straight d t end fraction equals 0  for t

table row cell 12 t minus 36 end cell equals 0 row cell 12 t end cell equals 36 row t equals 3 end table

So the minimum velocity occurs when t equals 3

(Note that, by the symmetry of quadratic graphs, that's halfway between the t values found in part (b)!)


Substitute t equals 3 into v open parentheses t close parentheses to find the velocity at that time


table row cell v open parentheses 3 close parentheses end cell equals cell 6 open parentheses 3 close parentheses squared minus 36 open parentheses 3 close parentheses plus 48 end cell row blank equals cell 54 minus 108 plus 48 end cell row blank equals cell negative 6 end cell end table 


Minimum velocity bold equals bold minus bold 6 bold space bold m bold divided by bold s 

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

      • Integrate velocity to find displacement

  • Similarly, since acceleration is the derivative of velocity (a equals fraction numerator straight d v over denominator straight d t end fraction)

    • space v equals integral a space straight d t

      • Integrate acceleration to find velocity

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

  • Without further information integration can only find s or v 'up to a constant of integration'

    • i.e.  integral v open parentheses t close parentheses space straight d t equals s open parentheses t close parentheses plus c

    • and  integral a open parentheses t close parentheses space straight d t equals v open parentheses t close parentheses plus c

  • To find the value of c we need additional information

    • Usually this will be the value of s or v at a particular time t 

  • Look out for the words “initial” or “initially”

    • This refers to time  t equals 0

  • Substitute the known values into your integration answers for s open parentheses t close parentheses or v open parentheses t close parentheses

    • and solve for c

Examiner Tips and Tricks

  • Read the question closely to spot any given values at particular times

    • These allow you to find the constant of integration

    • Remember that 'initially' means t equals 0

Worked Example

A particle P is moving along the x-axis.

At time t seconds (t greater or equal than 0) the velocity, v space straight m divided by straight s, of P is given by  v open parentheses t close parentheses equals 6 t squared minus t plus 3.

When t equals 0, the displacement of P from the origin is  negative 5 space straight m.

Find the displacement of P from the origin when  t equals 4.

Start by integrating the velocity to find an expression for the displacement

Note that this 'powers of t' integral works exactly the same as a 'powers of x' integral

Don't forget the constant of integration!



table row cell s open parentheses t close parentheses end cell equals cell integral open parentheses 6 t squared minus t plus 3 close parentheses space straight d t end cell row blank equals cell 6 open parentheses fraction numerator t to the power of 2 plus 1 end exponent over denominator 2 plus 1 end fraction close parentheses minus open parentheses fraction numerator t to the power of 1 plus 1 end exponent over denominator 1 plus 1 end fraction close parentheses plus 3 t plus c end cell row blank equals cell 2 t cubed minus 1 half t squared plus 3 t plus c end cell end table


We know that when t equals 0,  the displacement is  negative 5

Substitute these values in and solve for c

table row cell s open parentheses 0 close parentheses end cell equals cell negative 5 end cell row cell 2 open parentheses 0 close parentheses cubed minus 1 half open parentheses 0 close parentheses squared plus 3 open parentheses 0 close parentheses plus c end cell equals cell negative 5 end cell row cell 0 plus c end cell equals cell negative 5 end cell row c equals cell negative 5 end cell end table


Now we have a precise expression for s open parentheses t close parentheses

s open parentheses t close parentheses equals 2 t cubed minus 1 half t squared plus 3 t minus 5


Substitute in  t equals 4  to find the displacement at that time

table attributes columnalign right center left columnspacing 0px end attributes row cell s open parentheses 4 close parentheses end cell equals cell 2 open parentheses 4 close parentheses cubed minus 1 half open parentheses 4 close parentheses squared plus 3 open parentheses 4 close parentheses minus 5 end cell row blank equals cell 128 minus 8 plus 12 minus 5 end cell row blank equals 127 end table


Displacement bold equals bold 127 bold space bold m

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    Author
    Reviewer
    Roger B

    Author: Roger B

    Expertise: Maths Content Creator

    Roger's teaching experience stretches all the way back to 1992, and in that time he has taught students at all levels between Year 7 and university undergraduate. Having conducted and published postgraduate research into the mathematical theory behind quantum computing, he is more than confident in dealing with mathematics at any level the exam boards might throw at you.

    Dan Finlay

    Reviewer: Dan Finlay

    Expertise: Maths Lead

    Dan graduated from the University of Oxford with a First class degree in mathematics. As well as teaching maths for over 8 years, Dan has marked a range of exams for Edexcel, tutored students and taught A Level Accounting. Dan has a keen interest in statistics and probability and their real-life applications.