Compose & Resolve Velocities (Cambridge International IGCSE Further Maths)

Modelling Velocities with Vectors

How are velocities modelled by vectors?

• Although introduced as being about paths and distances between points, a vector can also represent a velocity

  • For example the velocity vector $( \begin{gathered} 3 \\ 7 \end{gathered} ) \mathrm{m} \mathrm{s}^{- 1}$ (or $( 3 i + 7 j ) \mathrm{m} \mathrm{s}^{- 1}$) would represent a particle or object moving

    • at $3ms^{-1}$ (metres per second) in the positive $x$ direction
    • and $7ms^{-1}$ in the positive $y$ direction

• Speed is different to velocity

  • Speed is a scalar quantity, velocity is a vector quantity

  • Velocity has direction, as well as magnitude

  • Speed is the magnitude of velocity

    • Therefore, speed can be found from velocity by finding its magnitude - i.e. using Pythagoras' theorem e.g. The speed of a particle travelling with velocity $( \begin{gathered} 6 \\ 8 \end{gathered} ) \mathrm{m} \mathrm{s}^{- 1}$ is $\sqrt{6^{2}+8^{2}}=10ms^{-1}$
    • An object travelling with velocity $( \begin{gathered} - 6 \\ - 8 \end{gathered} ) \mathrm{m} \mathrm{s}^{- 1}$ would have the same speed, $10ms^{-1}$ but be moving in the opposite direction

How do I find the position of a particle from a velocity vector?

• In general, problems refer to a particle's position vector, $( x i + y j ) \mathrm{m}$ (metres, from the origin) at time $t$ seconds after its motion has started

  • This position vector is often called $r$

• If the particle moves with (constant) velocity $vms^{-1}$, after $t$ seconds its position vector will be

  • $r=r_{0}+vt$ where $r_{0}$ is the initial position of the particle (i.e. its position at the start of the motion)

• For example if a particle starts at the point with coordinates $( 2 , - 3 )$ and moves with (constant) velocity $( 4 i + 7 j ) \mathrm{m} \mathrm{s}^{- 1}$ after 10 seconds its position will be

$\begin{gathered} r = ( 2 i - 3 j ) + 10 ( 4 i + 7 j ) \\ r = ( ( 2 + 40 ) i + ( - 3 + 70 ) j ) \\ r = ( 42 i + 67 j ) \mathrm{m} \end{gathered}$

How do I solve problems involving velocities and vectors?

• Solving problems involving velocity may involve using a variety of the skills covered in the vectors section

  • A resultant velocity may be comprised of two (or more) velocities

    • e.g. the velocity of a javelin will be influenced by the athlete's ability and the wind speed/direction
    • a resultant velocity is found by adding velocity vectors

• Problems may be phrased to distinguish the difference between speed and velocity

  • Speed is the magnitude of velocity (use Pythagoras' theorem)

• Problems may use position vectors

  • The initial position ($r_{0}$) is not necessarily the origin

    • $r=r_{0}+vt$

• There could be two particles to deal with in a problem

  • be clear about which particle has which velocity, position, etc

    • If two particles collide at time $t$ seconds, then (at time $t$) their position vectors will be equal
    • Two vectors are equal if their components are equal

Exam Hint

• Vector diagrams drawn previously to show paths and distances can still be used to visualise velocities

  • So use any given diagram, and if there isn't one, draw one!

• Read questions carefully - a common mistake is to give a final answer as a position vector when the question has asked for coordinates, or vice versa

  • e.g. A particle with position vector $( 3 i - 2 j )$ has coordinates $( 3 , - 2 )$

Worked Example A boat leaves a harbour (the origin) and sails with a (constant) resultant velocity comprising of the velocity produced by the boat's engine, $( i + 2 j ) \mathrm{m} \mathrm{s}^{- 1}$ and the velocity produced by the water current, $( 1.5 i + 2 j ) \mathrm{m} \mathrm{s}^{- 1}$.

a) Find the resultant velocity of the boat.

The resultant velocity will be the sum of the two velocities.

$v = ( i + 2 j ) + ( 1.5 i + 2 j ) v = ( 2.5 i + 4 j )$

Final answer: The resultant velocity of the boat is$(2.5i+4j)ms^{-1}$.

A second boat has position vector $( 27 i + 6 j ) \mathrm{m}$ at the same time as when the first boat leaves the harbour. The second boat sails with (constant) resultant velocity $( - 2 i + 3 j ) \mathrm{m} \mathrm{s}^{- 1}$.

b) Without intervention, the two boats will collide at time $t$ seconds.

i) Find the value of $t$.

ii) Find the coordinates of the point at which the boats will collide.

i) The position of a particle at time $t$ is $r=r_{0}+tv$ where $r_{0}$ is the initial position and $v$ is its (resultant) velocity.

The two boats will be in the same position when they collide.

$\therefore ( 0 i + 0 j ) + t ( 2.5 i + 4 j ) = ( 27 i + 6 j ) + t ( - 2 i + 3 j )$

Equating $i$ (or $j$) components gives an equation in $t$.

• *$\begin{gathered} 2.5 t & = 27 - 2 t \\ 4.5 t & = 27 \\ t = 6 \end{gathered}$ or $\begin{gathered} 4 t & = 6 + 3 t \\ t = 6 \end{gathered}$

Final answer: The boats collide at$t=6$seconds.

ii) Find the position of either boat at $t=6$ seconds.

$\begin{gathered} r = ( 0 i + 0 j ) + 6 ( 2.5 i + 4 j ) \\ r = ( 15 i + 24 j ) \end{gathered}$

The question asks for coordinates (rather than a position vector).

Final answer: The boats will collide at the point$(15,24)$.