Using Differentiation for Kinematics (Edexcel IGCSE Maths)

Kinematics

Spec point — 3.4E

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

What is a displacement function?

• The displacement of an object, $s$ metres, can be written as a function of time, $t$ seconds

  • $s = \mathrm{f} ( t )$

• Substitute a value of time in to find its displacement at that time

  • For example, $s=2t-t^{2}+1$

    • Initially, $t=0$ gives $s=2\times0-0^{2}+1=1$ (1 metre in front of the origin)
    • After 3 seconds, $t=3$ gives $s=2\times3-3^{2}+1=-2$ (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

  • $v=\frac{ds}{dt}$

• For example, if $s=t^{3}-4t^{2}+2t$ then $v=3t^{2}-8t+2$ (by differentiation)

  • The initial velocity ($t=0$) is $v=3\times0^{2}-8\times0+2=2$ ms^-1

  • The velocity after 1 second ($t=1$) is $v=3\times1^{2}-8\times1+2=-3$ 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

  • $a=\frac{dv}{dt}$

• For example, if $v=3t^{2}-8t+2$ then $a=6t-8$

  • 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

Exam Hint

• 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

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

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

Answer:

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

$\begin{gathered} s = 0^{3} - 6 \times 0^{2} - 3 \\ = - 3 \end{gathered}$

Distance is always positive, so convert -3 into 3

Final answer: 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.

Answer:

To find the velocity, differentiate the displacement

$v=\frac{ds}{dt}=3t^{2}-12t$

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

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

(c) Find how long, after $t=0$, it takes for the particle to come to rest.

Answer:

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 $t=4$

Final answer: 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.

Answer:

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

$a=\frac{dv}{dt}=6t-12$

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

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

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