NCEA Level 2KinematicsAS 91262

Kinematics: 7 exam questions solved step by step

What this lesson covers

  • Velocity is the derivative of displacement and acceleration is the derivative of velocity.
  • Going back up the chain needs anti-differentiation, and the starting values fix each constant.
Key idea

7 questions, one chain

  • s to v to a: differentiate.
  • a to v to s: anti-differentiate, + c.
  • At rest: v = 0. Greatest v: a = 0.
  • Units: m, m/s, m/s².
Your turn

Question 1 · Achieved

  • A cyclist's displacement after t seconds:
  • s = 4t² − t³ (metres)
  • Find v and a when t = 2.
Worked example

Q1: s = 4t² − t³

  • Differentiate to go down.
  • Units on every answer.
  1. s = 4t² − t³
  2. v = ds/dt = 8t − 3t²4t² → 8t, t³ → 3t²
  3. v(2) = 16 − 12 = 4 m/s
  4. a = dv/dt = 8 − 6tdifferentiate again
  5. a(2) = 8 − 12 = −4 m/s²
  6. v = 4 m/s, a = −4 m/s²
  7. v > 0, a < 0: slowing downinterpret
Your turn

Question 2 · Achieved

  • A particle has velocity v = 3t² + 2t (m/s).
  • It starts at s = 0.
  • Find its displacement after 4 seconds.
Worked example

Q2: v = 3t² + 2t

  • Going up: + c.
  • The start fixes c.
  1. s = t³ + t² + c3t² → t³, 2t → t²
  2. s(0) = 0 + 0 + c = 0starts at s = 0
  3. c = 0
  4. s = t³ + t²
  5. s(4) = 64 + 16 = 80
  6. 80 m from the start
  7. check: ds/dt = 3t² + 2t
Your turn

Question 3 · Merit

  • A particle moves along a line with
  • s = 2t³ − 15t² + 24t (metres)
  • When is it at rest, and where is it then?
Worked example

Q3: when is it at rest?

  • At rest: v = 0.
  • A negative s is allowed.
  1. v = 6t² − 30t + 24differentiate
  2. 6t² − 30t + 24 = 0at rest: v = 0
  3. 6(t² − 5t + 4) = 0take out 6
  4. 6(t − 1)(t − 4) = 0factorise
  5. t = 1 s or t = 4 s
  6. s(1) = 2 − 15 + 24 = 11 m
  7. s(4) = 128 − 240 + 96 = −16 m
  8. t = 1 s at 11 m; t = 4 s at −16 m
Your turn

Question 4 · Merit

  • An object has a = 6t − 4 (m/s²).
  • When t = 1, its velocity is 3 m/s.
  • Find v, then the velocity at t = 3.
  • Careful: the condition is at t = 1, not 0.
Worked example

Q4: a = 6t − 4

  • Write + c, then use t = 1.
  • c is not the 3 m/s.
  1. v = 6t²/2 − 4t + canti-differentiate
  2. = 3t² − 4t + c
  3. v(1) = 3 − 4 + c = 3v = 3 when t = 1
  4. c = 4
  5. v = 3t² − 4t + 4
  6. v(3) = 27 − 12 + 4 = 19 m/s
  7. check: dv/dt = 6t − 4
Your turn

Question 5 · Merit, in context

  • A cricket ball is hit straight up at 15 m/s.
  • It leaves the bat 1 m above the ground.
  • Take a = −10 m/s². Find its greatest height.
Worked example

Q5: form v and s

  • Up is positive.
  • Two constants: c, then k.
  1. a = −10
  2. v = −10t + canti-differentiate
  3. v(0) = 15, so c = 15hit up at 15 m/s
  4. v = 15 − 10t
  5. s = 15t − 5t² + kanti-differentiate
  6. s(0) = 1, so k = 11 m above the ground
  7. s = 15t − 5t² + 1
Worked example

Q5: the greatest height

  • Top of the flight: v = 0.
  • Answer in words, with units.
  1. v = 15 − 10t
  2. s = 15t − 5t² + 1
  3. 15 − 10t = 0top: v = 0
  4. t = 1.5 s
  5. v > 0 before, v < 0 afterso a maximum
  6. s(1.5) = 15 × 1.5 − 5 × 1.5² + 1
  7. = 22.5 − 11.25 + 1
  8. = 12.25 m
  9. greatest height 12.25 m
Your turn

Question 6 · Excellence

  • A particle starts at s = 0 with velocity
  • v = t² − 6t + 8 (m/s).
  • Find the total distance it travels
  • in the first 4 seconds.
Worked example

Q6: does it turn round?

  • Find s, with k.
  • Then find where v = 0.
  1. s = t³/3 − 3t² + 8t + kanti-differentiate
  2. s(0) = 0, so k = 0starts at s = 0
  3. t² − 6t + 8 = 0at rest: v = 0
  4. (t − 2)(t − 4) = 0factorise
  5. t = 2 s or t = 4 s
  6. v(1) = 3 > 0, v(3) = −1 < 0
  7. turns back at t = 2 s
Worked example

Q6: add the two legs

  • Split at the turn.
  • Distance is not displacement.
  1. s = t³/3 − 3t² + 8t
  2. s(2) = 8/3 − 12 + 16 = 20/3
  3. s(4) = 64/3 − 48 + 32 = 16/3
  4. 0 to 2 s: 20/3 m forward
  5. 2 to 4 s: 20/3 − 16/3 = 4/3 m back
  6. distance = 20/3 + 4/3 = 8 m
  7. displacement = 16/3 ≈ 5.33 mnot the distance
Your turn

Question 7 · Excellence, in context

  • A train leaves an Auckland station from rest.
  • a = 1 − t/12 (m/s²), until a = 0.
  • Find its greatest speed, and how far
  • it has gone by then.
Worked example

Q7: the greatest speed

  • From rest: v(0) = 0.
  • Greatest v: a = 0.
  1. v = t − t²/24 + canti-differentiate
  2. v(0) = 0, so c = 0from rest
  3. v = t − t²/24
  4. 1 − t/12 = 0greatest v: a = 0
  5. t = 12 s
  6. a > 0 before, a < 0 afterso v is a maximum
  7. v(12) = 12 − 144/24 = 6 m/s
Worked example

Q7: how far by then?

  • Up again: a new constant.
  • Answer in words.
  1. v = t − t²/24
  2. s = t²/2 − t³/72 + kanti-differentiate
  3. s(0) = 0, so k = 0starts at the station
  4. s(12) = 144/2 − 1728/72
  5. = 72 − 24 = 48 m
  6. greatest speed 6 m/s, at 48 m
  7. check: dv/dt = 1 − t/12
Summary

The method that solved all seven

  • Down by differentiating, up with + c, then interpret.
StepHow
Go downdifferentiate: s to v to a
Go upanti-differentiate, with + c
Constantssubstitute the given values
At restv = 0, then find s
Greatest va = 0, check the sign change
Distancesplit where v changes sign

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