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IBMathsDPApplications & Interpretation (AI)HLRevision Notes3. Geometry & TrigonometryModelling with VectorsKinematics with Vectors

Kinematics with Vectors (DP IB Applications & Interpretation (AI)): Revision Note

Amber

Written by: Amber

Reviewed by: Dan Finlay

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Kinematics using Vectors

What is kinematics?

  • Kinematics is the use of mathematics to model motion in objects

  • These are common terms used in kinematics:

    • Displacement describes the location of an object with respect to a fixed starting point

    • Velocity describes how the displacement changes over time

    • Acceleration describes how the velocity changes over time

  • A vector for the velocity describes the direction an object is moving in

  • The speed of an object is the magnitude of its velocity

How do I find whether two objects intersect?

  • For example, consider two objects with position vectors:

    • bold italic r subscript A equals open parentheses table row 7 row cell negative 1 end cell row cell negative 6 end cell end table close parentheses plus t open parentheses table row cell negative 2 end cell row 1 row 5 end table close parentheses

    • bold italic r subscript B equals open parentheses table row 1 row 5 row 4 end table close parentheses plus t open parentheses table row 1 row cell negative 2 end cell row 0 end table close parentheses

  • STEP 1
    Set the two equations equal to each other

    • open parentheses table row 7 row cell negative 1 end cell row cell negative 6 end cell end table close parentheses plus t open parentheses table row cell negative 2 end cell row 1 row 5 end table close parentheses equals open parentheses table row 1 row 5 row 4 end table close parentheses plus t open parentheses table row 1 row cell negative 2 end cell row 0 end table close parentheses

  • STEP2
    Form three equations using the components

    • 7 minus 2 t equals 1 plus t

    • negative 1 plus t equals 5 minus 2 t

    • negative 6 plus 5 t equals 4

  • STEP 3
    Solve any one of the equations

    • If the value is negative, then the objects do not intersect

      • negative 6 plus 5 t equals 4 rightwards double arrow t equals 2

  • STEP 4
    Check whether that value of t satisfies the other two equations

    • If it does, then the objects intersect

    • Otherwise, they do not

      • 7 minus 2 open parentheses 2 close parentheses equals 1 plus open parentheses 2 close parentheses equals 3

      • negative 1 plus open parentheses 2 close parentheses equals 5 minus 2 open parentheses 2 close parentheses equals 1

      • Therefore, they intersect

  • You can find the position vector of the point of intersection by substituting the value of t back into one of the position vectors of the objects

How do I find the shortest distance between two moving objects?

  • For example, consider two objects with position vectors:

    • bold italic r subscript A equals open parentheses table row 9 row 0 row cell negative 7 end cell end table close parentheses plus t open parentheses table row cell negative 2 end cell row 1 row 5 end table close parentheses

    • bold italic r subscript B equals open parentheses table row 0 row 6 row cell negative 1 end cell end table close parentheses plus t open parentheses table row 1 row cell negative 2 end cell row 0 end table close parentheses

  • STEP 1
    Find the displacement vector between the two position vectors

    • bold italic d equals open parentheses table row cell 9 minus 2 t end cell row t row cell negative 7 plus 5 t end cell end table close parentheses minus open parentheses table row t row cell 6 minus 2 t end cell row cell negative 1 end cell end table close parentheses equals open parentheses table row cell 9 minus 3 t end cell row cell negative 6 plus 3 t end cell row cell negative 6 plus 5 t end cell end table close parentheses

  • STEP 2
    Find the magnitude of this displacement vector

    • open vertical bar bold italic d close vertical bar equals square root of open parentheses 9 minus 3 t close parentheses squared plus open parentheses negative 6 plus 3 t close parentheses squared plus open parentheses negative 6 plus 5 t close parentheses squared end root

    • open vertical bar bold italic d close vertical bar equals square root of 43 t squared minus 150 t plus 153 end root

  • STEP 3
    Find the minimum value of the function under the square root

    • You can complete the square, use differentiation or use your GDC

      • Minimum of 43 t squared minus 150 t plus 153 is 954 over 43 and occurs when t equals 75 over 43

  • STEP 4
    Take the positive square root of this value

    • square root of 954 over 43 end root equals 4.710...

Worked Example

Two objects, A and B, are moving so that their position relative to a fixed point, O at time t, in minutes can be defined by the position vectors begin mathsize 16px style bold italic r subscript bold italic A blank equals blank open parentheses fraction numerator 3 over denominator negative 1 end fraction close parentheses plus t open parentheses fraction numerator negative 2 over denominator 4 end fraction close parentheses end style and begin mathsize 16px style bold italic r subscript bold italic B blank equals blank open parentheses 2 over 5 close parentheses plus t open parentheses fraction numerator 3 over denominator negative 1 end fraction close parentheses end style.

The unit vectors i and j are a displacement of 1 metre due East and North of O respectively.

a) Find the coordinates of the initial position of the two objects.

3-9-1-ib-ai-hl-kin-vectors-we-soltuion-a

b) Find the shortest distance between the two objects and the time at which this will occur.

3-9-1-ib-ai-hl-kin-vectors-we-soltuion-b

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    Author
    Reviewer
    Amber

    Author: Amber

    Expertise: Maths Content Creator

    Amber gained a first class degree in Mathematics & Meteorology from the University of Reading before training to become a teacher. She is passionate about teaching, having spent 8 years teaching GCSE and A Level Mathematics both in the UK and internationally. Amber loves creating bright and informative resources to help students reach their potential.

    Dan Finlay

    Reviewer: Dan Finlay

    Expertise: Maths Subject Lead

    Dan graduated from the University of Oxford with a First class degree in mathematics. As well as teaching maths for over 8 years, Dan has marked a range of exams for Edexcel, tutored students and taught A Level Accounting. Dan has a keen interest in statistics and probability and their real-life applications.