Mathematics · OCR H240 · Topic 3.02

Kinematics

Describing motion: displacement, velocity and acceleration, motion graphs, the suvat equations, calculus and projectiles.

  • 5 sub-topics
  • 2 in Year 12, 1 in Year 13
  • 84 practice questions
  • 5 animated lessons
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What the specification says

Every sub-topic OCR can examine in this topic, with its reference number from the official specification. The Y12 and Y13 tags show when each sub-topic is usually taught.

  1. 3.02aLanguage of kinematicsY12
  2. 3.02b-cGraphical representationY12
  3. 3.02d-eConstant accelerationY13
  4. 3.02f-gNon-uniform accelerationY13
  5. 3.02h-iGravity and projectilesY13

Try 3 questions

Easy, medium and hard questions from this topic, each with a worked solutions and the mark for every step. All 84 questions in this topic, free, no account needed.

Kinematics

Easy

Constant acceleration · Year 12 · 4 marks

A car accelerates uniformly from 4 m s−14\ \mathrm{m\,s^{-1}} to 20 m s−120\ \mathrm{m\,s^{-1}} in 8 seconds.
  1. (a)
    Find the acceleration of the car.
    (2)
  2. (b)
    Find the distance travelled in this time.
    (2)

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Part (a)2 marks

  1. Use v=u+atv = u + at with u=4u = 4, v=20v = 20, t=8t = 8.20=4+8a20 = 4 + 8a
  2. Solve for aa.a=2 m s−2a = 2\ \mathrm{m\,s^{-2}}

Part (b)2 marks

  1. Use s=12(u+v)ts = \tfrac{1}{2}(u + v)t, which needs no acceleration.s=12(4+20)×8s = \tfrac{1}{2}(4 + 20) \times 8
  2. Include units.s=96s = 96 m

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Kinematics

Medium

Thrown upwards · Year 12 · 4 marks

  1. A ball is thrown up at 14.7 m s⁻¹. Find the greatest height and the time until it returns to the same level.
    (4)
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  1. At the top, v=0v = 0.0=14.72−2(9.8)s0 = 14.7^2 - 2(9.8)s
  2. s=11.0s = 11.0 m
  3. Back at the start, s=0s = 0.0=14.7t−4.9t20 = 14.7t - 4.9t^2
  4. t=3t = 3 s

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Kinematics

Hard

Catching up · Year 12 · 5 marks

Car AA passes a junction at a constant 20 m s⁻¹. At that moment car BB starts from rest at the junction and accelerates at 2.5 m s⁻² in the same direction.
  1. Find when and where BB catches AA.
    (5)
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  1. Write each car's distance from the junction.sA=20ts_A = 20t, sB=1.25t2s_B = 1.25t^2
  2. Same distance means they are side by side (t=0t = 0 is the start).1.25t2=20t⇒t=161.25t^2 = 20t \Rightarrow t = 16 s
  3. 320320 m from the junction

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