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M11.3MechanicsStretch

Maxima and Minima Problems

Finding a maximum velocity or greatest height is an optimisation problem in disguise. The maximum or minimum of a quantity occurs where its rate of change is zero — so you differentiate and set the derivative to zero.

25 min Variable Acceleration
Edexcel Mechanics Y1 — Variable Acceleration playlist (Zeeshan Zamurred)Curated by Kepler, not produced by us. Watch it before the notes below.
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What you'll be able to do

  • Find maximum/minimum velocity
  • Find greatest height or displacement
  • Use the condition: rate of change = 0
  • Interpret the result in context
1

The condition for a maximum

A quantity is at a maximum or minimum where its derivative (rate of change) is zero. So maximum velocity\textbf{maximum velocity} occurs when dvdt=0\dfrac{dv}{dt} = 0 — that is, when the acceleration is zero.

max/min velocity⇒dvdt=a=0\text{max/min velocity} \Rightarrow \dfrac{dv}{dt} = a = 0
Velocity is greatest when acceleration is zero.
Example 1A particle has v=12t−3t2v = 12t - 3t^2. Find the maximum velocity.
1Acceleration a=dvdt=12−6t=0a = \dfrac{dv}{dt} = 12 - 6t = 0, so t=2t = 2.
2Max velocity: v=12(2)−3(2)2=24−12v = 12(2) - 3(2)^2 = 24 - 12.
Answer1212 m/s at t=2t = 2 s

Tip — Maximum velocity ⟺ acceleration = 0. Maximum displacement ⟺ velocity = 0.

2

Greatest displacement

The greatest displacement (e.g. maximum height) occurs when the velocity\textbf{velocity} is zero — the object stops moving in that direction before turning back. Set v=0v = 0, find tt, then substitute into ss.

3

Method summary

Differentiate the relevant quantity, set the derivative to zero to find the critical time, then substitute back to get the maximum or minimum value. State the answer with its time and units.

Formula recap

dvdt=0⇒max/min velocity\dfrac{dv}{dt} = 0 \Rightarrow \text{max/min velocity}
Acceleration zero.
v=0⇒greatest displacementv = 0 \Rightarrow \text{greatest displacement}
Turning point of motion.
differentiate→=0→substitute back\text{differentiate} \to =0 \to \text{substitute back}
The method.

Common mistakes to avoid

Setting velocity to zero to find maximum velocity.
Maximum velocity is where acceleration (dv/dt) = 0; v = 0 gives greatest displacement.
Finding the time but not the actual maximum value.
Substitute the critical time back into the quantity for its value.

Key takeaways

  • Max/min velocity: set acceleration (dv/dt) = 0.
  • Greatest displacement: set velocity (v) = 0.
  • Differentiate, set to zero, then substitute back for the value.

Test yourself

Answer a few questions, then review anything you missed.

Next step

Check your understanding

Work through questions on Maxima and Minima Problems. Use the feedback to decide what needs another look.

4 past-paper questions · 14 marks

Practise this topic

Related lessons

All 5 lessons

Put the methods into practice: see how A-Level maths duels work. An account is required to play; the overview and sample are public.