Differentiate displacement to interpret variable motion
Use Paper 1 derivatives to find velocity and acceleration, identify motion intervals and decide when speed increases.
Exam scope and assessment notes
Mechanics (Paper 4, AS & A Level) · Mathematics 9709, 2026–2027, section 4.2. Motion is restricted to one dimension. Choose a positive direction; distance and speed are scalars, while displacement, velocity and acceleration carry a direction. Deceleration means decreasing speed, not necessarily negative acceleration. Calculus uses Paper 1 techniques only. Constant-acceleration formulae require constant acceleration on the stated interval; use a new interval and its initial velocity when conditions change. Where vertical motion occurs, use the supplied g = 10 m s⁻² and the stated no-air-resistance model. Written practice is self-checked; human subject-teacher review remains pending.
Before you begin
- Differentiate powers and factor a quadratic.
- Build a sign table over a specified interval.
Work on paper. Try each question before revealing help, and explain why your method works.
Quick readiness check
If s = 2t³ − 3t² + 4, what is v?
What you will learn
- Differentiate displacement twice with respect to time.
- Use roots and signs to interpret rest, reversal and changing speed.
Differentiate with respect to the actual time variable
If displacement s is given as a differentiable function of time, velocity is v = ds/dt and acceleration is a = dv/dt = d²s/dt². The units change from m to m s⁻¹ to m s⁻². For s = t³ − 6t² + 9t + 2 on 0 ≤ t ≤ 4, differentiation gives v = 3t² − 12t + 9 = 3(t − 1)(t − 3), then a = 6t − 12. The constant initial position disappears from derivatives but still matters when evaluating positions.
Solving v = 0 gives t = 1 and t = 3 s. A sign table shows positive velocity before 1, negative velocity between 1 and 3, and positive velocity after 3. Thus the particle reverses at both roots. Evaluate s at the roots to mark the displacement graph: s(0) = 2, s(1) = 6, s(3) = 2 and s(4) = 6 m. Solving a = 0 instead gives t = 2 s, when velocity is −3 rather than zero.
Interpret velocity and acceleration together
For this cubic example, speed decreases on 0 < t < 1 because v > 0 and a < 0. It increases on 1 < t < 2 because both are negative, decreases on 2 < t < 3 because v < 0 and a > 0, and increases on 3 < t < 4 because both are positive. The velocity graph is a parabola with minimum (2, −3); its time-axis crossings determine reversal and its tangent slopes determine acceleration.
Do not use constant-acceleration formulae for this whole interval: a = 6t − 12 varies with time. Differentiation gives instantaneous information even when speed changes non-uniformly. A root of velocity can be a momentary rest without reversal if velocity does not change sign, so inspect signs on both sides. At a non-zero velocity, va > 0 means increasing speed and va < 0 means decreasing speed; discuss zero cases separately.
Pause and explain
For v = 3(t − 1)(t − 3), a = 6t − 12, what happens on 1 < t < 2?
Worked example
For s = t³ − 6t² + 9t + 2 m on 0 ≤ t ≤ 4 s, find v, a, rest times and the time when acceleration is zero.
Show the worked solution
- Differentiate to obtain v = 3t² − 12t + 9 = 3(t − 1)(t − 3).
- Differentiate again to obtain a = 6t − 12, so a = 0 at 2 s.
- Rest occurs at 1 and 3 s, with velocity changing sign at each; at 2 s the velocity is −3 m s⁻¹.
Answer v = 3(t − 1)(t − 3); a = 6t − 12; reversals at 1 and 3 s; zero acceleration at 2 s with v = −3 m s⁻¹.
Try a topic test
Choose Easy, Normal or Difficult. Save your answers and open worked review after finishing.
Test this topic →From guided practice to a new situation
Write your answer and reasoning first. Use a hint only if you are stuck. The model solution is for self-checking; your written work is not automatically marked.
For s = 2t³ − 3t² + 4 m, calculate velocity and acceleration at t = 2 s.
Give me a hint
Differentiate once for velocity and twice for acceleration before substituting.
Compare my reasoning
- Velocity is v = 6t² − 6t.
- Acceleration is a = 12t − 6.
- At t = 2, v = 24 − 12 = 12 m s⁻¹ and a = 24 − 6 = 18 m s⁻².
Velocity +12 m s⁻¹; acceleration +18 m s⁻².
Look for these in your work
- Use the power rule on both non-constant terms.
- State the different velocity and acceleration units.
For s = 5 − 4t + t³ m, calculate v and a at t = 2 s and decide whether speed increases at that instant.
Give me a hint
Compare the signs after evaluating both derivatives.
Compare my reasoning
- Velocity is v = −4 + 3t², giving v(2) = 8 m s⁻¹.
- Acceleration is a = 6t, giving a(2) = 12 m s⁻².
- Both are positive, so the particle moves positively and its speed increases.
v = +8 m s⁻¹; a = +12 m s⁻²; speed increasing.
Look for these in your work
- Differentiate the linear term to its constant rate.
- Base the speed interpretation on both signs.
A particle has v = −6 + 3t m s⁻¹ for 0 ≤ t ≤ 4. Find acceleration and rest time, and compare its speed changes at t = 1 and t = 3.
Give me a hint
One velocity function can describe opposite motion directions at different times.
Compare my reasoning
- Differentiation gives constant a = +3 m s⁻², and v = 0 at t = 2 s.
- At 1 s, v = −3 m s⁻¹, so positive acceleration decreases speed.
- At 3 s, v = +3 m s⁻¹, so the same acceleration increases speed.
a = +3 m s⁻²; reversal at 2 s; speed decreasing at 1 s and increasing at 3 s.
Look for these in your work
- Distinguish acceleration from the changing velocity.
- Explain why a constant acceleration can first slow and then speed the particle.
Common mistakes
- Differentiate with respect to time rather than treating displacement as the independent variable.
- A zero acceleration is not necessarily a zero velocity.
- Inspect velocity signs before declaring a reversal.
What can you explain now?
If v = (t − 2)² on 0 ≤ t ≤ 4, does motion reverse at t = 2?
Compare with the explanation
No; velocity is non-negative on both sides.
The particle is momentarily at rest but resumes motion in the same positive direction.
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Make it stick
Differentiate s to v to a, solve for the relevant zero and use interval signs to interpret motion and speed.
If you needed the solution, close it and solve the problem again from a blank page. A correct answer is stronger when you can explain the reason for each step.