Calculus for Kinematics (Edexcel IGCSE Further Maths)

Differentiation for Kinematics

Spec point — 9C

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

    • $v=\frac{ds}{dt}$ (differentiate displacement to get velocity)

  • acceleration is the rate of change of velocity

    • $a=\frac{dv}{dt}$ (differentiate velocity to get acceleration)

  • so acceleration is also the second derivative of displacement

    • $a=\frac{d^{2}s}{dt^{2}}$ (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 $\frac{dv}{dt}=0$

Worked Example The displacement, $s$ metres, of a particle at time $t$ seconds, is modelled by $s(t)=2t^{3}-18t^{2}+48t$, $t\geq0$.

(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!

$\begin{gathered} v = \frac{ \mathrm{d} s}{ \mathrm{d} t} \\ = 2 ( 3 t^{3 - 1} ) - 18 ( 2 t^{2 - 1} ) + 48 \end{gathered}$

$v(t)=6t^{2}-36t+48$

Now differentiate the velocity to find the acceleration.

$\begin{array}{rcl} a = \frac{ \mathrm{d} v}{ \mathrm{d} t} \\ = 6 ( 2 t^{2 - 1} ) - 36 \end{array}$

$a(t)=12t-36$

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

'Instantaneously at rest' means that the velocity is zero

So solve $v ( t ) = 0$ for $t$

$\begin{array}{rcl} 6 t^{2} - 36 t + 48 & = 0 \\ 6 ( t^{2} - 6 t + 8 ) & = 0 \\ t^{2} - 6 t + 8 & = 0 \\ ( t - 2 ) ( t - 4 ) & = 0 \end{array}$ $t=2\mathrm{or}t=4$

Final answer: The particle is at rest at$t=2$ seconds and at $t=4$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 $v ( t ) = 6 t^{2} - 36 t + 48$ is a 'u-shaped' parabola

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

That minimum point will occur when $\frac{dv}{dt}=0$

So start by solving $a=\frac{dv}{dt}=0$ for $t$

$\begin{gathered} 12 t - 36 & = 0 \\ 12 t & = 36 \\ t = 3 \end{gathered}$

So the minimum velocity occurs when $t=3$

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

Substitute $t=3$ into $v ( t )$ to find the velocity at that time

$\begin{gathered} v ( 3 ) & = 6 ( 3 )^{2} - 36 ( 3 ) + 48 \\ = 54 - 108 + 48 \\ = - 6 \end{gathered}$* *

Final answer: Minimum velocity$=-6m/s$

Integration for Kinematics

Spec point — 9C

How is integration used in kinematics?

• Since velocity is the derivative of displacement ($v=\frac{ds}{dt}$) it follows that

  • $s = \int v \mathrm{d} t$

    • Integrate velocity to find displacement

• Similarly, since acceleration is the derivative of velocity($a=\frac{dv}{dt}$)

  • $v = \int a \mathrm{d} t$

    • 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. $\int v ( t ) \mathrm{d} t = s ( t ) + c$
  • and $\int a ( t ) \mathrm{d} t = v ( t ) + c$

• 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 $t=0$

• Substitute the known values into your integration answers for $s ( t )$ or $v ( t )$

  • and solve for $c$

Exam Hint

• 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 $t=0$

Worked Example A particle $P$ is moving along the $x$-axis.

At time $t$ seconds ($t\geq0$) the velocity, $vm/s$, of $P$ is given by $v ( t ) = 6 t^{2} - t + 3$.

When $t=0$, the displacement of $P$ from the origin is $-5m$.

Find the displacement of $P$ from the origin when $t=4$.

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!

$\begin{gathered} s ( t ) & = \int ( 6 t^{2} - t + 3 ) \mathrm{d} t \\ = 6 ( \frac{t^{2 + 1} }{2 + 1} ) - ( \frac{t^{1 + 1} }{1 + 1} ) + 3 t + c \\ = 2 t^{3} - 1 \frac{1}{2} t^{2} + 3 t + c \end{gathered}$

We know that when $t=0$, the displacement is $-5$

Substitute these values in and solve for $c$

$\begin{gathered} s ( 0 ) & = - 5 \\ 2 ( 0 )^{3} - 1 \frac{1}{2} ( 0 )^{2} + 3 ( 0 ) + c & = - 5 \\ 0 + c & = - 5 \\ c = - 5 \end{gathered}$

Now we have a precise expression for $s ( t )$

$s ( t ) = 2 t^{3} - 1 \frac{1}{2} t^{2} + 3 t - 5$

Substitute in $t=4$ to find the displacement at that time

$\begin{array}{rcl} s ( 4 ) & = 2 ( 4 )^{3} - 1 \frac{1}{2} ( 4 )^{2} + 3 ( 4 ) - 5 \\ = 128 - 8 + 12 - 5 \\ = 127 \end{array}$

Final answer: Displacement$=127m$