A-Level Maths 9709 / Topic 4.2 / AS & A Level

Differentiate down, integrate up.

When acceleration changes with time the suvat equations fail. Calculus from Pure 1 takes over: differentiate displacement to get velocity and acceleration, integrate the other way and use the starting values to fix the constant.

Topic 4.2All kinematics of motion in a straight line lessons →

The key idea

\(v = \dfrac{ds}{dt}\) and \(a = \dfrac{dv}{dt}\). Going back, \(v = \displaystyle\int a\,dt\) and \(s = \displaystyle\int v\,dt\), each with a constant found from a known value (usually at \(t = 0\)). Distance travelled is the total area between the velocity graph and the axis, so split the integral wherever \(v = 0\).

Fig. 1: v = t² − 4t + 3 for 0 ≤ t ≤ 4 (to scale). The particle reverses at t = 1 and t = 3. Displacement = 4/3 − 4/3 + 4/3 = 4/3 m, but distance = 4/3 + 4/3 + 4/3 = 4 m

Section 01

Differentiate to go down.

Given displacement as a function of time, differentiate once for velocity and again for acceleration. "At rest" (or "instantaneously at rest") means \(v = 0\); "maximum velocity" usually means \(a = 0\).

Given	To find	Do
\(s\)	\(v\), then \(a\)	differentiate
\(a\)	\(v\), then \(s\)	integrate, adding a constant each time
\(v\)	distance	integrate, split where \(v = 0\)

Worked example Times at rest

A particle moves in a straight line with displacement \(s = t^3 - 6t^2 + 9t\) metres from O at time \(t\) s. Find when it is at rest, and its acceleration at those times.

\(v = \dfrac{ds}{dt} = 3t^2 - 12t + 9 = 3(t - 1)(t - 3)\).
\(v = 0\) when \(t = 1\) and \(t = 3\).
\(a = \dfrac{dv}{dt} = 6t - 12\): at \(t = 1\), \(a = -6\); at \(t = 3\), \(a = 6\).

Answer: At rest at \(t = 1\) (\(a = -6\ \text{m s}^{-2}\)) and \(t = 3\) (\(a = 6\ \text{m s}^{-2}\)).

Worked example Distance from a displacement formula

For the same particle, find the total distance travelled in the first 4 s.

It changes direction at \(t = 1\) and \(t = 3\), so find \(s\) at the turning points.
\(s(0) = 0\), \(s(1) = 4\), \(s(3) = 0\), \(s(4) = 64 - 96 + 36 = 4\).
Distance \(= |4 - 0| + |0 - 4| + |4 - 0| = 4 + 4 + 4\).

Answer: 12 m (although the displacement at \(t = 4\) is only 4 m).

Examiner trap

\(s(4) - s(0)\) is the DISPLACEMENT. If the particle turns round in the interval, the distance is bigger: find every time when \(v = 0\) first.

Section 02

Integrate to go up.

Integrating loses information, so each integral carries a constant. Use a given condition, such as "starts from rest at O" (\(v = 0\) and \(s = 0\) when \(t = 0\)), to find it.

Worked example From acceleration to displacement

A particle starts at O with velocity \(3\ \text{m s}^{-1}\). Its acceleration is \(a = 4 - 2t\). Find its maximum velocity, and its displacement from O when \(t = 3\).

\(v = \displaystyle\int (4 - 2t)\,dt = 4t - t^2 + c\); \(v = 3\) at \(t = 0\), so \(c = 3\): \(v = 3 + 4t - t^2\).
Maximum \(v\) when \(a = 0\): \(t = 2\), so \(v = 3 + 8 - 4 = 7\ \text{m s}^{-1}\).
\(s = \displaystyle\int v\,dt = 3t + 2t^2 - \tfrac{1}{3}t^3 + k\); \(s = 0\) at \(t = 0\), so \(k = 0\).
\(s(3) = 9 + 18 - 9 = 18\) m.

Answer: Maximum velocity \(7\ \text{m s}^{-1}\); displacement 18 m.

Your turn

A particle starts from O with velocity \(v = 3\sqrt{t}\ \text{m s}^{-1}\). Find its acceleration when \(t = 4\) and its distance from O at that time.

Examiner trap

Forgetting \(+c\). If the particle starts with \(v = 3\), dropping the constant gives the wrong velocity at every later time and every answer after it.

Stage 1 · Learn

Check your understanding

Four quick checks. Each correct answer earns XP and records this skill on your profile.

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How is velocity obtained from displacement \(s\) as a function of \(t\)?

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A particle has \(s = t^3\). Find its velocity at \(t = 2\).

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A particle has \(v = 5t^2 - 2t\). Find its acceleration at \(t = 1\).

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Why must you split \(\displaystyle\int v\,dt\) where \(v = 0\) to find distance?

Section 03

Distance when the velocity changes sign.

The integral \(\displaystyle\int_a^b v\,dt\) gives displacement, because area below the axis counts as negative. For distance, split at every zero of \(v\) and add the sizes of the pieces, as in Fig. 1.

Worked example Splitting the integral

A particle has velocity \(v = t^2 - 4t + 3\ \text{m s}^{-1}\). Find the distance it travels from \(t = 0\) to \(t = 4\).

\(v = (t - 1)(t - 3) = 0\) at \(t = 1\) and \(t = 3\).
With \(S(t) = \tfrac{1}{3}t^3 - 2t^2 + 3t\): \(S(0) = 0\), \(S(1) = \tfrac{4}{3}\), \(S(3) = 0\), \(S(4) = \tfrac{4}{3}\).
Pieces: \(\tfrac{4}{3}\), \(-\tfrac{4}{3}\), \(\tfrac{4}{3}\).
Distance \(= \tfrac{4}{3} + \tfrac{4}{3} + \tfrac{4}{3} = 4\) m.

Answer: 4 m (the displacement is only \(\tfrac{4}{3}\) m).

Examiner trap

Integrating straight from 0 to 4 gives \(\tfrac{4}{3}\), because the negative middle piece cancels one positive piece. A sketch of \(v\) against \(t\) shows you where to split.

Stage 2 · Exam

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A particle starts with velocity \(3\ \text{m s}^{-1}\) and has acceleration \(a = 4 - 2t\). Find its maximum velocity.

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A particle has velocity \(v = t^2 - 4t + 3\ \text{m s}^{-1}\). Find the distance travelled from \(t = 0\) to \(t = 4\).

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A particle starts from O with velocity \(2\ \text{m s}^{-1}\) and acceleration \(a = 6t - 4\). Find its displacement from O when \(t = 3\).

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A particle has displacement \(s = t^3 - 6t^2 + 9t\) m. Find the distance travelled in the first 4 s.

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Six original Cambridge-style questions

Differentiating a displacement, a square-root velocity, maximum velocity from an acceleration, times at rest, a particle that returns to its start, and a two-stage journey.