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Compose & Resolve Velocities (Cambridge (CIE) O Level Additional Maths): Revision Note

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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 open parentheses table row 3 row 7 end table close parentheses space straight m space straight s to the power of negative 1 end exponent (or open parentheses 3 bold i plus 7 bold j close parentheses space straight m space straight s to the power of negative 1 end exponent) would represent a particle or object moving

      • at 3 space straight m space straight s to the power of negative 1 end exponent (metres per second) in the positive x direction

      • and 7 space straight m space straight s to the power of negative 1 space end exponent 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 open parentheses table row 6 row 8 end table close parentheses space straight m space straight s to the power of negative 1 end exponent is square root of 6 squared plus 8 squared end root equals 10 space straight m space straight s to the power of negative 1 end exponent

      • An object travelling with velocity open parentheses table row cell negative 6 end cell row cell negative 8 end cell end table close parentheses space straight m space straight s to the power of negative 1 end exponent would have the same speed, 10 space straight m space straight s to the power of negative 1 end exponent 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, open parentheses x bold i plus y bold j close parentheses space straight m (metres, from the origin) at time t seconds after its motion has started

    • This position vector is often called bold r

  • If the particle moves with (constant) velocity bold v space straight m space straight s to the power of negative 1 end exponent, after t seconds its position vector will be

    • bold r equals bold r subscript bold 0 plus bold v t where bold r subscript bold 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 open parentheses 2 comma space minus 3 close parentheses and moves with (constant) velocity open parentheses 4 bold i plus 7 bold j close parentheses space straight m space straight s to the power of negative 1 end exponent after 10 seconds its position will be

table row bold r equals cell open parentheses 2 bold i minus 3 bold j close parentheses plus 10 open parentheses 4 bold i plus 7 bold j close parentheses end cell row bold r equals cell open parentheses open parentheses 2 plus 40 close parentheses bold i plus open parentheses negative 3 plus 70 close parentheses bold j close parentheses end cell row bold r equals cell open parentheses 42 bold i plus 67 bold j close parentheses space straight m end cell end table

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 (bold r subscript bold 0) is not necessarily the origin

      • bold r equals bold r subscript bold 0 plus bold v t

  • 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

Examiner Tips and Tricks

  • 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 open parentheses 3 bold i minus 2 bold j close parentheses has coordinates open parentheses 3 comma space minus 2 close parentheses

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, open parentheses bold i plus 2 bold j close parentheses space straight m space straight s to the power of negative 1 end exponent and the velocity produced by the water current, open parentheses 1.5 bold i plus 2 bold j close parentheses space straight m space straight s to the power of negative 1 end exponent.

a) Find the resultant velocity of the boat.

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

bold v equals open parentheses bold i plus 2 bold j close parentheses plus open parentheses 1.5 bold i plus 2 bold j close parentheses bold v equals open parentheses 2.5 bold i plus 4 bold j close parentheses

The resultant velocity of the boat is .

A second boat has position vector open parentheses 27 bold i plus 6 bold j close parentheses space straight m at the same time as when the first boat leaves the harbour.
The second boat sails with (constant) resultant velocity open parentheses negative 2 bold i plus 3 bold j close parentheses space straight m space straight s to the power of negative 1 end exponent.

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 bold r equals bold r subscript bold 0 plus t bold v where r subscript 0 is the initial position and bold v is its (resultant) velocity.

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

therefore space open parentheses 0 bold i plus 0 bold j close parentheses plus t open parentheses 2.5 bold i plus 4 bold j close parentheses equals open parentheses 27 bold i plus 6 bold j close parentheses plus t open parentheses negative 2 bold i plus 3 bold j close parentheses

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

table row cell 2.5 t end cell equals cell 27 minus 2 t end cell row cell 4.5 t end cell equals 27 row t equals 6 end table      or       table row cell 4 t end cell equals cell 6 plus 3 t end cell row t equals 6 end table

The boats collide at bold italic t bold equals bold 6 seconds.

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

table row bold r equals cell open parentheses 0 bold i plus 0 bold j close parentheses plus 6 open parentheses 2.5 bold i plus 4 bold j close parentheses end cell row bold r equals cell open parentheses 15 bold i plus 24 bold j close parentheses end cell end table

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

The boats will collide at the point stretchy left parenthesis 15 comma space 24 stretchy right parenthesis.

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