Calculus: Kinematics and Practical Problems
Differentiation tells you how fast something is changing. Edexcel uses it for moving objects (displacement → velocity → acceleration) and for finding the best value in practical problems, such as the largest area or the lowest cost. These questions are worth 4–6 marks and follow a very predictable pattern.
1Quick recap: differentiating
The ruleIf y = axⁿ, then dydx = naxⁿ⁻¹. Differentiate each term separately; a constant term differentiates to 0.
This works for negative powers too: y = 50x = 50x⁻¹ gives dydx = −50x⁻² = −50x².
For full notes on the power rule and stationary points, see Differentiation.
2Displacement, velocity and acceleration
For an object moving in a straight line, with time t:
| Quantity | Symbol | How to get it |
|---|---|---|
| Displacement | s | given |
| Velocity | v | v = dsdt |
| Acceleration | a | a = dvdt |
Key ideas → v → a: differentiate to move right along the chain.
Worked example
A particle moves so that its displacement after t seconds is s = t³ − 6t² + 9t metres. Find its velocity and acceleration when t = 2.
- v = dsdt = 3t² − 12t + 9. At t = 2: v = 12 − 24 + 9 = −3 m/s.
- a = dvdt = 6t − 12. At t = 2: a = 12 − 12 = 0 m/s².
Exam tipA negative velocity means the object is moving backwards, in the negative direction. It does not mean it is slowing down.
3When is the object at rest?
“At rest” or “instantaneously at rest” means v = 0. Solve the velocity equation.
Worked example
For s = t³ − 6t² + 9t, find when the particle is at rest and its displacement at those times.
- v = 3t² − 12t + 9 = 0 → t² − 4t + 3 = 0 → (t − 1)(t − 3) = 0.
- t = 1: s = 1 − 6 + 9 = 4 m.
- t = 3: s = 27 − 54 + 27 = 0 m (back at the start).
4Maximum height and greatest distance
A maximum or minimum of s happens when ds/dt = 0, which is when v = 0. A maximum or minimum of v happens when dv/dt = 0, which is when a = 0.
Worked example
A ball is thrown upwards. Its height after t seconds is h = 20t − 5t² metres. Find its maximum height.
- dhdt = 20 − 10t = 0 → t = 2.
- h = 20(2) − 5(4) = 20.
5Practical problems: finding the best value
The method is always the same:
- Write the quantity to maximise or minimise (area, volume, cost, …) as a formula in one variable. Questions often give or guide this.
- Differentiate and set the derivative equal to 0.
- Solve for the variable, rejecting values that make no sense in context.
- Substitute back to find the maximum or minimum value.
- Show it is a maximum or a minimum (see Section 6).
Worked example
A farmer has 40 m of fencing to make a rectangular pen against a long wall. The wall forms one side. The two sides at right angles to the wall are x m long. Find the largest possible area.
- The side parallel to the wall is 40 − 2x, so A = x(40 − 2x) = 40x − 2x².
- dAdx = 40 − 4x = 0 → x = 10.
- A = 10 × 20 = 200.
Worked example
An open box is made from a 24 cm square sheet by cutting a square of side x cm from each corner and folding up the sides. Find the maximum volume.
- V = x(24 − 2x)².
- dVdx = (24 − 2x)² − 4x(24 − 2x) = (24 − 2x)(24 − 6x).
- = 0 when x = 12 (no box: V = 0) or x = 4.
- V = 4 × 16² = 1024.
6Is it a maximum or a minimum?
Edexcel accepts either method below. Always show one.
Method 1Gradient either side. Find dy/dx just before and just after the stationary point.
+ then − → maximum
− then + → minimum
+ then − → maximum
− then + → minimum
Method 2Shape of the graph. A negative x² (or t²) term gives a ∩ shape, so a quadratic’s stationary point is a maximum. For a cubic, use Method 1.
Worked example
Show that y = x³ − 12x + 5 has a minimum at x = 2.
- dydx = 3x² − 12. At x = 2: 12 − 12 = 0, so it is a stationary point.
- At x = 1: 3 − 12 = −9 (negative). At x = 3: 27 − 12 = 15 (positive).
- The gradient goes from − to +, so it is a minimum. y = 8 − 24 + 5 = −11.
Common mistakes
- Differentiating v to get s, or integrating. At IGCSE you only ever differentiate: s → v → a.
- Thinking “at rest” means s = 0. It means v = 0.
- Forgetting to substitute back. Questions usually ask for the maximum area, not just the value of x.
- Not rejecting impossible solutions, such as x = 12 giving a box with no base.
- Losing the minus sign when differentiating negative powers: (50x⁻¹)′ = −50x⁻².
★Key Vocabulary Recap
- Displacement s
- Distance from a fixed point in a given direction.
- Velocity v
- Rate of change of displacement, ds/dt.
- Acceleration a
- Rate of change of velocity, dv/dt.
- Stationary point
- A point where the gradient (derivative) is zero.
- Optimise
- Find the maximum or minimum value of a quantity.
Specification reference (Pearson Edexcel 4XMA1, Unit 2 Higher)
- Understand the concept of a variable rate of change; differentiate integer powers of x.
- Determine gradients, rates of change, stationary points, turning points (maxima and minima) by differentiation and relate these to graphs.
- Distinguish between maxima and minima by considering the general shape of the graph only.
- Apply calculus to linear kinematics and to other simple practical problems.
Next stepReady to practise? Try the 12 exam-style questions, then check your working against the worked solutions.
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