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Cambridge IGCSE · Topic 14.14
IGCSE Add Maths 0606 / Topic 14 / IGCSE

Differentiate to slow down, integrate to travel.

A particle moving along a straight line has a displacement \(s\), a velocity \(v\) and an acceleration \(a\), and calculus links all three. Differentiate to go from \(s\) to \(v\) to \(a\); integrate, with a constant fixed by the starting conditions, to come back. The same ideas read every motion graph: the gradient of one is the next, the area under one is the last.

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The key idea

For motion in a straight line, \(v = \dfrac{\mathrm{d}s}{\mathrm{d}t}\) and \(a = \dfrac{\mathrm{d}v}{\mathrm{d}t}\); going back, \(v = \displaystyle\int a\,\mathrm{d}t\) and \(s = \displaystyle\int v\,\mathrm{d}t\), each with a constant found from a known value (usually at \(t = 0\)). The particle is at rest when \(v = 0\) and changes direction when \(v\) changes sign. Displacement is \(\displaystyle\int_{t_1}^{t_2} v\,\mathrm{d}t\); total distance adds the sizes of the pieces between the times when \(v = 0\). Speed is \(|v|\).

1 2 3 5 4 −6 6 0 t v v = 8 + 2t − t² area above = 80/3 area below = 10/3 v < 0: moving back displacement = 80/3 − 10/3 = 70/3 m distance = 80/3 + 10/3 = 30 m (split at t = 4, where v = 0)
Fig. 1: the velocity-time graph of v = 8 + 2t - t² for 0 ≤ t ≤ 5. The particle moves forwards (v above the axis) until it is at rest at t = 4, then moves backwards. The area above the axis is 80/3 and the area below is 10/3. Displacement counts the area below as negative, 70/3 m; distance adds the two sizes, 30 m
Section 01

Differentiate down the chain.

Take a fixed point \(O\) on the line. The displacement \(s\) of a particle from \(O\) is measured along the line, positive on one side and negative on the other. Its velocity is the rate of change of displacement and its acceleration is the rate of change of velocity, so each comes from the one before by differentiating with respect to time \(t\).

\[ s \;\xrightarrow{\ \frac{\mathrm{d}}{\mathrm{d}t}\ }\; v = \frac{\mathrm{d}s}{\mathrm{d}t} \;\xrightarrow{\ \frac{\mathrm{d}}{\mathrm{d}t}\ }\; a = \frac{\mathrm{d}v}{\mathrm{d}t} = \frac{\mathrm{d}^{2}s}{\mathrm{d}t^{2}} \]

Worked example Acceleration from velocity
A particle moves in a straight line so that \(t\) seconds after passing a fixed point \(O\) its velocity is \(v = 2t^{2} - 16t + 30\) m s-1. Find (a) the acceleration when \(t = 2\) (b) the times when the particle is at rest.
  1. (a) \(a = \dfrac{\mathrm{d}v}{\mathrm{d}t} = 4t - 16\). When \(t = 2\): \(a = 8 - 16 = -8\).
  2. The negative sign means the velocity is decreasing at that moment.
  3. (b) At rest: \(2t^{2} - 16t + 30 = 0\), so \(t^{2} - 8t + 15 = 0\) and \((t - 3)(t - 5) = 0\).
  4. \(v\) is positive before \(t = 3\), negative between 3 and 5, positive after 5, so the particle changes direction at both times.
Answer: (a) \(-8\) m s-2   (b) \(t = 3\) and \(t = 5\)
Worked example From displacement to both
A particle moves so that its displacement from \(O\) after \(t\) seconds is \(s = t^{3} - 9t^{2} + 24t\) metres. Find the displacement and the acceleration at each time when the particle is at rest.
  1. \(v = \dfrac{\mathrm{d}s}{\mathrm{d}t} = 3t^{2} - 18t + 24 = 3(t - 2)(t - 4)\), so the particle is at rest when \(t = 2\) and \(t = 4\).
  2. \(a = \dfrac{\mathrm{d}v}{\mathrm{d}t} = 6t - 18\).
  3. \(t = 2\): \(s = 8 - 36 + 48 = 20\) and \(a = -6\).
  4. \(t = 4\): \(s = 64 - 144 + 96 = 16\) and \(a = 6\).
Answer: \(t = 2\): \(s = 20\) m, \(a = -6\) m s-2; \(t = 4\): \(s = 16\) m, \(a = 6\) m s-2
Your turn
\(s = 2t^{3} - 15t^{2} + 36t\). Find the times when the particle is at rest and the acceleration at each.
Examiner trap

'At rest' means \(v = 0\), NOT \(a = 0\). A ball thrown upwards is at rest at the top while its acceleration is still \(-9.8\). Setting \(a = 0\) finds the time of greatest or least VELOCITY, a different question.

Section 02

Integrate back up.

Going the other way, integrate: \(v = \int a\,\mathrm{d}t\) and \(s = \int v\,\mathrm{d}t\). Each integration brings a constant, and the question always gives you a value to fix it, such as 'starts from rest' (\(v = 0\) when \(t = 0\)) or 'passes through \(O\) when \(t = 0\)' (\(s = 0\) when \(t = 0\)). The same rules apply to the trig and exponential models in this course; remember that \(t\) is in radians.

Worked example Polynomial acceleration
A particle starts from rest at \(O\) and its acceleration after \(t\) seconds is \(a = 6t - 4\) m s-2. Find (a) \(v\) in terms of \(t\) (b) \(s\) in terms of \(t\) (c) the time when the particle is next at \(O\).
  1. (a) \(v = \int (6t - 4)\,\mathrm{d}t = 3t^{2} - 4t + c\). From rest: \(v = 0\) when \(t = 0\), so \(c = 0\) and \(v = 3t^{2} - 4t\).
  2. (b) \(s = \int (3t^{2} - 4t)\,\mathrm{d}t = t^{3} - 2t^{2} + k\). At \(O\) when \(t = 0\): \(k = 0\), so \(s = t^{3} - 2t^{2}\).
  3. (c) \(s = 0\): \(t^{2}(t - 2) = 0\), so \(t = 0\) (the start) or \(t = 2\).
  4. Check: \(s(3) = 27 - 18 = 9\), so by \(t = 3\) it has passed \(O\) and is 9 m beyond it, as expected.
Answer: (a) \(v = 3t^{2} - 4t\)   (b) \(s = t^{3} - 2t^{2}\)   (c) \(t = 2\) s
Worked example A trig velocity
A particle moves so that \(v = 4\sin 2t\) m s-1, and \(s = 0\) when \(t = 0\). Find \(s\) and \(a\) in terms of \(t\), and the displacement when the particle first comes to rest after \(t = 0\).
  1. \(s = \int 4\sin 2t\,\mathrm{d}t = -2\cos 2t + c\). At \(t = 0\): \(0 = -2 + c\), so \(c = 2\) and \(s = 2 - 2\cos 2t\).
  2. \(a = \dfrac{\mathrm{d}v}{\mathrm{d}t} = 8\cos 2t\).
  3. At rest: \(\sin 2t = 0\), so \(2t = \pi\) is the first positive solution and \(t = \dfrac{\pi}{2}\).
  4. \(s = 2 - 2\cos \pi = 2 + 2 = 4\).
Answer: \(s = 2 - 2\cos 2t\), \(a = 8\cos 2t\); at \(t = \dfrac{\pi}{2}\), \(s = 4\) m
Your turn
A particle has displacement \(s = 5 - 5e^{-t}\) metres. Find its initial velocity, and the time and displacement when \(v = 1\).
Examiner trap

Never drop the constant. Without it, \(a = 6t - 4\) would give the same \(v\) for a particle that starts from rest and one that starts at 10 m s-1. And find \(c\) from the condition given: \(s = 0\) at \(t = 0\) does not make \(c = 0\) when the integral has a cosine term, since \(\cos 0 = 1\).

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A particle has displacement \(s = t^{3} - 4t^{2} + 5t\) metres. Find its velocity when \(t = 2\).

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A particle is instantaneously at rest when

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A particle has acceleration \(a = 6t\) m s-2, and \(v = 2\) when \(t = 0\). Find \(v\) in terms of \(t\).

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On a velocity-time graph, what does the gradient give?

Section 03

Distance, speed and the graphs.

The integral \(\int_{t_1}^{t_2} v\,\mathrm{d}t\) is the change in displacement. When the particle turns round, the backward part counts as negative and cancels some of the forward part. For the total DISTANCE, find the times when \(v = 0\) inside the interval, integrate over each piece separately, and add the sizes (Fig. 1). Speed is \(|v|\), so the speed-time graph is the velocity-time graph with the parts below the axis reflected above it.

GraphIts gradient isThe area under it is
displacement-time (\(s\)-\(t\))velocity \(v\)(not used)
velocity-time (\(v\)-\(t\))acceleration \(a\)displacement (below the axis counts negative)
acceleration-time (\(a\)-\(t\))(not used)change in velocity
distance-time and speed-timespeeddistance (speed-time graph)

Turning points of the \(s\)-\(t\) graph are where \(v = 0\); turning points of the \(v\)-\(t\) graph are where \(a = 0\). The distance-time graph never goes down: after each turning point of \(s\) it carries on rising.

Worked example Displacement against distance
A particle moves with velocity \(v = 8 + 2t - t^{2}\) m s-1 for \(0 \le t \le 5\). Find its displacement and the total distance it travels.
  1. At rest: \(t^{2} - 2t - 8 = 0\), so \((t - 4)(t + 2) = 0\) and \(t = 4\) (reject \(-2\)). The particle turns round at \(t = 4\).
  2. \(\displaystyle\int_{0}^{4} (8 + 2t - t^{2})\,\mathrm{d}t = \left[8t + t^{2} - \frac{t^{3}}{3}\right]_{0}^{4} = 32 + 16 - \frac{64}{3} = \frac{80}{3}\).
  3. \(\displaystyle\int_{4}^{5} (8 + 2t - t^{2})\,\mathrm{d}t = \left(40 + 25 - \frac{125}{3}\right) - \frac{80}{3} = \frac{70}{3} - \frac{80}{3} = -\frac{10}{3}\).
  4. Displacement \(= \dfrac{80}{3} - \dfrac{10}{3} = \dfrac{70}{3}\); distance \(= \dfrac{80}{3} + \dfrac{10}{3} = 30\).
Answer: displacement \(\dfrac{70}{3} \approx 23.3\) m; distance \(30\) m
Worked example Reading the four graphs
A particle has displacement \(s = t^{3} - 6t^{2} + 9t\) metres for \(0 \le t \le 4\). Describe the \(s\)-\(t\), distance-time, speed-time and \(a\)-\(t\) graphs, and find the total distance travelled.
  1. \(v = 3t^{2} - 12t + 9 = 3(t - 1)(t - 3)\) and \(a = 6t - 12\). Values: \(s(0) = 0\), \(s(1) = 4\), \(s(3) = 0\), \(s(4) = 4\).
  2. \(s\)-\(t\): a cubic rising from 0 to a maximum of 4 at \(t = 1\), falling to a minimum of 0 at \(t = 3\), rising to 4 at \(t = 4\).
  3. Distance-time: rises to 4 at \(t = 1\), keeps rising to \(4 + 4 = 8\) at \(t = 3\), then to \(12\) at \(t = 4\). It never falls.
  4. Speed-time: \(|v|\) starts at 9, falls to 0 at \(t = 1\), rises to 3 at \(t = 2\) (where \(v = -3\) is least), falls to 0 at \(t = 3\), then rises to 9 at \(t = 4\).
  5. \(a\)-\(t\): a straight line from \(-12\) at \(t = 0\) to \(12\) at \(t = 4\), crossing zero at \(t = 2\), where the \(v\)-\(t\) graph has its turning point.
Answer: total distance \(4 + 4 + 4 = 12\) m (displacement only \(4\) m)
Your turn
A particle has constant acceleration and velocity \(v = 6 - 2t\) m s-1 for \(0 \le t \le 5\). Find its displacement and the distance travelled.
Examiner trap

Integrating \(v\) straight from the start time to the end time gives the DISPLACEMENT, not the distance: any backward motion cancels. Always solve \(v = 0\) first, split the integral at every root inside the interval, and add the sizes.

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A particle moves in a straight line with velocity \(v = 2t^{2} - 14t + 20\) m s-1. Find its acceleration when it is first at rest.

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A particle moves with velocity \(v = 6e^{-2t}\) m s-1. Find the distance it travels between \(t = 0\) and \(t = \ln 2\).

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A particle moves with velocity \(v = t^{2} - 6t + 5\) m s-1. Find the total distance it travels in the first 3 seconds.

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A particle has acceleration \(a = 4 - 2t\) m s-2 and its initial velocity is 5 m s-1. Find its maximum velocity.

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Six original Cambridge-style questions
Differentiating displacement and velocity, integrating with initial conditions, polynomial, trig and exponential models, total distance against displacement, and a full motion problem with its graphs.