Calculus for Kinematics | Edexcel IGCSE Further Maths Revision Notes 2019

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
  - $v = \frac{d s}{d t}$ (differentiate displacement to get velocity)
- acceleration is the rate of change of velocity
  - $a = \frac{d v}{d t}$ (differentiate velocity to get acceleration)
- so acceleration is also the second derivative of displacement
  - $a = \frac{d^{2} s}{d t^{2}}$ (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 $\frac{d v}{d t} = 0$

Worked example

The displacement, $s$ metres, of a particle at time $t$ seconds, is modelled by $s (t) = 2 t^{3} - 18 t^{2} + 48 t$ , $t \geq 0$ .

(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!

\begin{array}{rcl} v & = & \frac{d s}{d t} \\ = & 2 (3 t^{3 - 1}) - 18 (2 t^{2 - 1}) + 48 \end{array}

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

Now differentiate the velocity to find the acceleration.

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

a (t) = 12 t - 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 or t = 4

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{d v}{d t} = 0

So start by solving

a = \frac{d v}{d t} = 0

for

t

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

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{array}{rcl} v (3) & = & 6 {(3)}^{2} - 36 (3) + 48 \\ = & 54 - 108 + 48 \\ = & - 6 \end{array}

Minimum velocity $= - 6 m / s$

Integration for Kinematics

How is integration used in kinematics?

Since velocity is the derivative of displacement ( $v = \frac{d s}{d t}$ ) it follows that
- - Integrate velocity to find displacement

Similarly, since acceleration is the derivative of velocity ( $a = \frac{d v}{d t}$ )
- $v = \int a d t$
  - Integrate acceleration to find velocity

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

Without further information integration can only find or 'up to a constant of integration'
- i.e. $\int v (t) d t = s (t) + c$
- and $\int a (t) d t = v (t) + c$
To find the value of 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 Tip

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 ( $t \geq 0$ ) the velocity, $v m / s$ , of $P$ is given by $v (t) = 6 t^{2} - t + 3$ .

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

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!

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

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

Substitute these values in and solve for $c$

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

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

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

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

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

Displacement $= 127 m$

Calculus for Kinematics (Edexcel IGCSE Further Maths)

Revision Note

Differentiation for Kinematics

How is differentiation used in kinematics?

Worked example

Integration for Kinematics

How is integration used in kinematics?

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

Exam Tip

Worked example

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Author: Roger