Using Differentiation for Kinematics (Edexcel IGCSE Maths A (Modular))

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Written by: Mark Curtis

Reviewed by: Dan Finlay

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Kinematics

What does kinematics mean?

Kinematics is the study of the motion of an object
- It is a branch of Physics
Objects are called particles
- They are modelled as single moving points
Over time, the particles move and
- can be at different distances from a fixed origin (displacement)
- can move with different speeds in different directions (velocity)
- can speed up or slow down (acceleration)

What is displacement?

The displacement of an object is how far away it is from a fixed origin
- It can be positive (in front of the origin)
- or negative (behind the origin)
Do not confuse displacement with distance
- Distance is always positive!
- Displacement can have a $\pm$ sign
Displacement is given the letter $s$ in kinematics
- Do not confuse this letter for speed!
- Displacement is measured in metres

Diagrams showing displacement examples: one particle 4 metres right of point 0, another 5 metres left of point 0. Text explains displacement in each case.

What is a displacement function?

The displacement of an object, $s$ metres, can be written as a function of time, $t$ seconds
- $s = f (t)$
Substitute a value of time in to find its displacement at that time
- For example, $s = 2 t - t^{2} + 1$
  - Initially, $t = 0$ gives $s = 2 \times 0 - 0^{2} + 1 = 1$ (1 metre in front of the origin)
  - After 3 seconds, $t = 3$ gives $s = 2 \times 3 - 3^{2} + 1 = - 2$ (2 metres behind the origin)

What is the velocity and how do I find it?

Velocity is the speed and direction of an object
- It is positive if moving forwards
- It is negative if moving backwards
- Do not confuse velocity and speed
  - Speed is always positive!
To find the velocity of an object, $v$ metres per second, differentiate its displacement function
- $v = \frac{d s}{d t}$
For example, if $s = t^{3} - 4 t^{2} + 2 t$ then $v = 3 t^{2} - 8 t + 2$ (by differentiation)
- The initial velocity ( $t = 0$ ) is $v = 3 \times 0^{2} - 8 \times 0 + 2 = 2$ ms^-1
- The velocity after 1 second ( $t = 1$ ) is $v = 3 \times 1^{2} - 8 \times 1 + 2 = - 3$ ms^-1
  - Its speed is 3 ms^-1
If a velocity is zero at any point in time, it is said to be at instantaneous rest
- It is stationary (not moving) at that instant in time
  - but not stationary all the time
- To find the times at which the particle is at rest, set $v = 0$ and solve to find $t$

What is the acceleration and how do I find it?

Acceleration is rate at which the velocity changes
- It is positive if speeding up (when moving forwards)
- It is negative if slowing down (when moving forwards)
  - A negative acceleration is also called a deceleration
  - The magnitude of acceleration is always positive
To find the acceleration of an object, $a$ metres per second per second, differentiate its velocity function
- $a = \frac{d v}{d t}$
For example, if $v = 3 t^{2} - 8 t + 2$ then $a = 6 t - 8$
- You can substitute times in to find accelerations
It the acceleration is always zero then the particle moves at a constant speed

How do I find the acceleration from the displacement?

You differentiate displacement to get velocity, then differentiate velocity to get acceleration
- So you differentiate displacement twice to get acceleration

Diagram showing the relationship between displacement (s), velocity (v), and acceleration (a), with differentiation denoted as ds/dt for velocity and dv/dt for acceleration.

Examiner Tips and Tricks

Harder exam questions may jump back and forth between displacement, velocity and acceleration
- so make sure you use the labels $s = . . .$ , $v = . . .$ and $a = . . .$ to make your working clear

Worked Example

A particle moves along a straight line.

The displacement of the particle from a fixed point, O, on the line at time $t$ seconds is $s$ metres, where

$s = t^{3} - 6 t^{2} - 3$

(a) Find the initial distance of the particle from O.

Initial means $t = 0$
Substitute $t = 0$ into $s$ to find the initial displacement

$\begin{array}{rcl} s & = & 0^{3} - 6 \times 0^{2} - 3 \\ = & - 3 \end{array}$

Distance is always positive, so convert -3 into 3

The particle is initially at a distance of 3 metres from O

(b) Find an expression for the velocity, $v$ ms^-1, at time $t$ seconds.

To find the velocity, differentiate the displacement

$v = \frac{d s}{d t} = 3 t^{2} - 12 t$

This is an expression for the velocity in terms of time, $t$

$v = 3 t^{2} - 12 t$ ms^-1

The particle is at rest when $v = 0$
Set $v = 0$ and solve to find $t$

$0 = 3 t^{2} - 12 t 0 = t^{2} - 4 t 0 = t (t - 4)$

$t = 0$ or $t = 4$

After $t = 0$ the next point of rest is $t = 4$

After $t = 0$ , it takes 4 seconds for the particle to come to rest

(d) Find the time at which the particle is decelerating at 3 ms^-2.

A deceleration of 3 means an acceleration of -3
Differentiate the velocity function to find acceleration

$a = \frac{d v}{d t} = 6 t - 12$

Set $a = - 3$ and solve for $t$

$6 t - 12 = - 3 6 t = 9 t = 1.5$

The particle is decelerating at 3 ms^-2 at 1.5 seconds

Last updated: 18 October 2024

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