NCEA Level 2KinematicsAS 91262

Kinematics: displacement, velocity and acceleration

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

One chain, three quantities

  • Displacement s: where it is.
  • Velocity v: how fast s changes.
  • Acceleration a: how fast v changes.
SymbolMeaningUnits
sdisplacementmetres
v = ds/dtvelocitym/s
a = dv/dtaccelerationm/s²
Worked example

Down the chain: differentiate

  • Differentiate to go down.
  • Factorise v for later.
  1. s = t³ − 6t² + 9t
  2. v = ds/dt = 3t² − 12t + 9differentiate
  3. = 3(t² − 4t + 3)take out 3
  4. = 3(t − 1)(t − 3)factorise
  5. a = dv/dt = 6t − 12differentiate again
Worked example

When is it at rest?

  • At rest: v = 0.
  • Substitute into s and a.
  1. v = 3(t − 1)(t − 3)
  2. 3(t − 1)(t − 3) = 0at rest: v = 0
  3. t = 1 s or t = 3 s
  4. s(1) = 1 − 6 + 9 = 4 m
  5. a(1) = 6 − 12 = −6 m/s²
  6. s(3) = 27 − 54 + 27 = 0 mback at the start
Worked example

Velocity is the gradient of s

  • Rising graph: moving forward.
  • Flat at t = 1 and t = 3: at rest.
  • Falling in between: moving back.
Velocity is the gradient of s
Worked example

Up the chain: anti-differentiate

  • Each step up needs + c.
  • The start value fixes it.
  1. a = 6 − 2t
  2. v = 6t − 2t²/2 + canti-differentiate
  3. = 6t − t² + c
  4. v(0) = 0 − 0 + c = 5starts at 5 m/s
  5. c = 5
  6. v = 6t − t² + 5
Worked example

Up again: a new constant

  • A new constant, k.
  • It starts at s = 0.
  1. v = 6t − t² + 5
  2. s = 6t²/2 − t³/3 + 5t + kanti-differentiate
  3. = 3t² − t³/3 + 5t + k
  4. s(0) = 0 − 0 + 0 + k = 0starts at s = 0
  5. k = 0
  6. s = 3t² − t³/3 + 5t
  7. check: ds/dt = 6t − t² + 5
Key idea

Show how you found c

  • Write + c, then substitute t = 0.
  • Say what the starting value was.
  • The examiner wants this line for Merit.
Worked example

The maximum velocity

  • Max velocity: a = 0.
  • Then substitute t = 3.
  1. a = 6 − 2t = 0max v: a = 0
  2. t = 3 s
  3. da/dt = −2 < 0a maximum
  4. v(3) = 6(3) − 3² + 5
  5. = 18 − 9 + 5 = 14 m/s
  6. s(3) = 3(3²) − 3³/3 + 5(3)
  7. = 27 − 9 + 15 = 33 m
Key idea

Read the question words

  • "At rest": v = 0.
  • "Starts from rest": v = 0 when t = 0.
  • "Constant speed": a = 0.
Worked example

Starting from rest at the origin

  • From rest: v(0) = 0.
  • At the origin: s(0) = 0.
  1. a = 6t
  2. v = 6t²/2 + c = 3t² + canti-differentiate
  3. v(0) = 0 + c = 0, c = 0from rest
  4. s = 3t³/3 + k = t³ + kanti-differentiate
  5. s(0) = 0: k = 0, s = t³at the origin
  6. v(2) = 3 × 2² = 12 m/s
  7. s(2) = 2³ = 8 m
Key idea

Where the marks are

  • Achieved: differentiate or anti-differentiate once.
  • Merit: find the constants, with working.
  • Excellence: a full chain, answered in context.
Summary

What to take away

  • Down by differentiating, up with + c.
Down the chaindifferentiate: s to v to a
Up the chainanti-differentiate, with + c
Starting valuesfix each constant
At restv = 0
Max or min va = 0
Unitsm, m/s, m/s²
Your turn

Your turn: four questions

  • Pause here and answer each one.
  • Questions 3 and 4 need + c.
  • Give units.
Question
1s = 2t³ − 3t². Find v and a.
2When is that object at rest?
3a = 4 m/s² and v = 2 m/s at t = 0. Find v.
4It starts at s = 0. How far has it gone after 3 s?
Show answer

1. v = 6t² − 6t, a = 12t − 6. 2. 6t(t − 1) = 0, so t = 0 and t = 1 second. 3. v = 4t + c; v(0) = 2, so v = 4t + 2. 4. s = 2t² + 2t + k, k = 0, so s(3) = 18 + 6 = 24 m.

Worked example

The answers: questions 1 and 2

  1. 1. s = 2t³ − 3t²
  2. v = ds/dt = 6t² − 6tdifferentiate
  3. a = dv/dt = 12t − 6differentiate again
  4. 2. 6t² − 6t = 0at rest: v = 0
  5. 6t(t − 1) = 0take out 6t
  6. t = 0 s or t = 1 s
Worked example

The answers: questions 3 and 4

  1. 3. v = 4t + canti-differentiate a = 4
  2. v(0) = 0 + c = 2, so c = 22 m/s at the start
  3. v = 4t + 2
  4. 4. s = 4t²/2 + 2t + kanti-differentiate v
  5. = 2t² + 2t + k
  6. s(0) = 0 + 0 + k = 0, so k = 0starts at s = 0
  7. s(3) = 2 × 3² + 2 × 3
  8. = 18 + 6 = 24 m

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