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Rearranging formulaeIB MYP Maths Extended: Subtopic test

10 questions, 27 marks

IB MYP Maths Extended

Rearranging formulae

Total 27 marks

Name

Class

Date

  1. 1
    A car accelerates in a straight line with constant acceleration aa m/s². It starts at speed uu m/s and has speed vv m/s after tt seconds, where v=u+atv=u+at.
    (a)
    Make aa the subject of the formula.
    [1 mark]
    • Aa=v−uta=\frac{v-u}{t}
    • Ba=v+uta=\frac{v+u}{t}
    • Ca=vt−ua=\frac{v}{t}-u
    • Da=(v−u)ta=(v-u)t
    (b)
    Make uu the subject of the formula.
    [1 mark]
    • Au=at−vu=at-v
    • Bu=v+atu=v+at
    • Cu=v−atu=v-at
    • Du=v−atu=\frac{v-a}{t}
    (c)
    A car starts at 77 m/s and reaches 2525 m/s with acceleration 33 m/s². Rearrange the formula to make tt the subject and find tt.
    [2 marks]

    Total for question 1: 4 marks

  2. 2
    The area AA of a trapezium with parallel sides aa and bb and perpendicular height hh is A=12(a+b)hA=\frac{1}{2}(a+b)h.
    (a)
    Make hh the subject of the formula.
    [1 mark]
    • Ah=A2(a+b)h=\frac{A}{2(a+b)}
    • Bh=2Aa+bh=\frac{2A}{a+b}
    • Ch=2A−(a+b)h=2A-(a+b)
    • Dh=A(a+b)2h=\frac{A(a+b)}{2}
    (b)
    Make aa the subject of the formula.
    [1 mark]
    • Aa=2Ah+ba=\frac{2A}{h}+b
    • Ba=2A−bha=\frac{2A-b}{h}
    • Ca=A2h−ba=\frac{A}{2h}-b
    • Da=2Ah−ba=\frac{2A}{h}-b
    (c)
    A trapezium has area 4444 cm2^2, height 88 cm and one parallel side a=7a=7 cm. Make bb the subject of the formula and find bb.
    [2 marks]

    Total for question 2: 4 marks

  3. 3
    The volume VV of a cone with base radius rr and height hh is V=13πr2hV=\frac{1}{3}\pi r^2h.
    (a)
    Make rr the subject of the formula.
    [3 marks]
    (b)
    A cone has height 88 cm and volume 96π96\pi cm3^3. Find its radius. A second cone has the same radius and volume 192π192\pi cm3^3. Find the height of the second cone.
    [4 marks]

    Total for question 3: 7 marks

  4. 4
    A stone is dropped from a bridge. The distance dd metres it has fallen after tt seconds is modelled by d=5t2d=5t^2, ignoring air resistance.
    (a)
    (i) Make tt the subject of the formula.
    (ii) The bridge is
    8080 m above a river. Find the time the stone takes to reach the river.
    (iii) Show that a stone that falls for twice as long as another stone falls four times as far.
    [6 marks]
    (b)
    On another planet the model is d=12gt2d=\frac{1}{2}gt^2, where gg m/s² is the acceleration due to gravity.
    (i) Make
    gg the subject of the formula.
    (ii) A stone falls
    2020 m in 2.52.5 s on this planet. Find gg.
    (iii) On Earth the model
    d=5t2d=5t^2 corresponds to g=10g=10. Justify which planet has the weaker gravity, and give one limitation of the model.
    [6 marks]

    Total for question 4: 12 marks

End of questions

Written by the Exaim team, led by Shaun Daswani (Head of Upper Secondary, Improve ME Institute; MSc Financial Mathematics, Imperial College London; BSc, UCL) and Jason Daswani (operational lead, Improve ME Institute; LSE).