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Topic 14 · 14.14–14.15

Calculus III: kinematics

Syllabus 14.14–14.15 · Papers 1 and 2

🎯What you need to be able to do

  • Differentiate displacement to get velocity and acceleration, and integrate the other way using given initial conditions.
  • Interpret “at rest”, “changes direction”, “returns to \(O\)”, “initially” and “maximum velocity”.
  • Tell displacement from distance, and velocity from speed.
  • Draw and use displacement–time, distance–time, velocity–time, speed–time and acceleration–time graphs.

📚The mathematics

The chain of derivatives

\[ s \;\xrightarrow{\ \frac{d}{dt}\ }\; v = \frac{ds}{dt} \;\xrightarrow{\ \frac{d}{dt}\ }\; a = \frac{dv}{dt} = \frac{d^{2}s}{dt^{2}} \]

Going back up the chain is integration, and each step brings a constant that you find from the given conditions (“starts from rest at \(O\)” means \( v = 0 \) and \( s = 0 \) when \( t = 0 \)). Acceleration may be variable or constant; the method is the same.

What the words mean

  • Initially: at \( t = 0 \).
  • At (instantaneous) rest: \( v = 0 \).
  • Changes direction: \(v\) changes sign — usually where \( v = 0 \), but check it actually changes sign there.
  • Returns to \(O\): \( s = 0 \) again.
  • Maximum or minimum velocity: \( a = \dfrac{dv}{dt} = 0 \).
  • Speed is \( |v| \); distance is the total length travelled, never negative.

Distance versus displacement

Displacement is where the particle is relative to \(O\). Distance is how far it has gone. They differ as soon as the particle turns round. To find the distance travelled over an interval: find every time the particle is at rest inside it, work out \(s\) at the start, at each turning time and at the end, and add up the sizes of the changes.

Three stacked graphs for s = t cubed minus 9 t squared plus 24 t over 0 to 5 seconds. Top: displacement rises to 20 at t = 2, falls to 16 at t = 4 and returns to 20 at t = 5, with the distance travelled drawn dashed, rising to 28. Middle: the velocity 3(t minus 2)(t minus 4) dips below zero between 2 and 4, and the speed is its reflection above the axis. Bottom: acceleration 6t minus 18, a straight line crossing zero at t = 3.
The worked example’s five graphs. Where \(v\) is negative (\( 2 < t < 4 \)), displacement falls but distance keeps rising, and speed is the reflection of velocity.

Reading the graphs

  • The gradient of a displacement–time graph is velocity; of a velocity–time graph, acceleration.
  • The area under a velocity–time graph is displacement (area below the axis counts negative); under a speed–time graph it is distance.
  • The speed–time graph is the velocity–time graph with the negative parts reflected, exactly like \( y = |f(x)| \) in Topic 1; distance–time never decreases.

✏️Worked example 1 (no calculator)

A particle moves in a straight line so that, \(t\) seconds after passing a fixed point \(O\), its displacement from \(O\) is \( s = t^{3} - 9t^{2} + 24t \) metres, for \( 0 \le t \le 5 \). (a) Find the times at which the particle is at rest. [3] (b) Find the acceleration when \( t = 1 \). [2] (c) Find the total distance travelled in the 5 seconds. [3]

(a) \( v = 3t^{2} - 18t + 24 = 3(t - 2)(t - 4) \), so the particle is at rest at \( t = 2 \) and \( t = 4 \).

(b) \( a = 6t - 18 \), so \( a = -12 \) m s−2 at \( t = 1 \). (The particle is still moving forwards, but slowing down.)

(c) \( s(0) = 0 \), \( s(2) = 20 \), \( s(4) = 16 \), \( s(5) = 20 \). The particle goes out 20 m, comes back 4 m, then goes forward 4 m again:

\[ \text{distance} = 20 + 4 + 4 = 28 \text{ m}, \qquad \text{displacement at } t = 5 \text{ is } 20 \text{ m} \]
Check it. Integrate the speed: \( \int_{0}^{2} v\,dt = 20 \), \( \int_{2}^{4} v\,dt = -4 \), \( \int_{4}^{5} v\,dt = 4 \). The sizes add to 28 ✓, and the signed values add to 20, the displacement ✓.
Answering “distance” with \( s(5) - s(0) \). That gives 20, the displacement. Whenever a question says distance, look for times of rest inside the interval first.

✏️Worked example 2 (calculator)

A particle starts from rest at \(O\) and moves in a straight line with velocity \( v = 8 - 8e^{-0.5t} \) m s−1. (a) Find the acceleration when \( t = 2 \). [2] (b) Find the displacement from \(O\) when \( t = 4 \). [4] (c) State the value that \(v\) approaches as \(t\) increases. [1]

(a) \( a = \dfrac{dv}{dt} = 4e^{-0.5t} \); at \( t = 2 \), \( a = 4e^{-1} = 1.47 \) m s−2.

(b) \( s = \int (8 - 8e^{-0.5t})\,dt = 8t + 16e^{-0.5t} + c \). At \( t = 0 \), \( s = 0 \): \( 0 = 16 + c \), so \( c = -16 \). At \( t = 4 \): \( s = 32 + 16e^{-2} - 16 = 18.2 \) m.

(c) \( e^{-0.5t} \to 0 \), so \( v \to 8 \) m s−1. Since \( v > 0 \) for all \( t > 0 \), the particle never turns round, and here the distance equals the displacement.

Assuming \( c = 0 \). \( 8t + 16e^{-0.5t} \) is 16 at \( t = 0 \), not 0. The exponential term does not vanish at the start, so the constant is almost never zero in these questions.

📝Practise

Written in the style of the current Paper 1 (no calculator) and Paper 2 (calculator) questions.

1. (No calculator.) A particle has velocity \( v = 2t^{2} - 14t + 20 \) m s−1. Find the times when it is at rest, and its acceleration when \( t = 1 \). [4]
\( 2(t^{2} - 7t + 10) = 2(t - 2)(t - 5) = 0 \Rightarrow t = 2, 5 \). \( a = 4t - 14 = -10 \) m s−2 at \( t = 1 \).
2. (No calculator.) A particle passes \(O\) at \( t = 0 \) with velocity 3 m s−1, and its acceleration is \( a = 6t - 4 \). (a) Find \(v\) and \(s\) in terms of \(t\). (b) Show that the particle is never at rest. [6]
(a) \( v = 3t^{2} - 4t + c \), with \( v = 3 \) at \( t = 0 \): \( v = 3t^{2} - 4t + 3 \). Then \( s = t^{3} - 2t^{2} + 3t \) (the constant is 0 because \( s = 0 \) at \( t = 0 \)). (b) \( v = 0 \) would need \( 3t^{2} - 4t + 3 = 0 \), whose discriminant is \( 16 - 36 < 0 \): no real roots, so \( v \) is never zero.
3. (No calculator.) A particle moves so that \( s = 2\sin 3t \) m. Find the first time after \( t = 0 \) at which it is at rest, and its acceleration at that time. [4]
\( v = 6\cos 3t = 0 \) first when \( 3t = \tfrac{\pi}{2} \), \( t = \tfrac{\pi}{6} \). \( a = -18\sin 3t = -18\sin\tfrac{\pi}{2} = -18 \) m s−2.
4. (Calculator.) A particle has velocity \( v = 5e^{-0.2t} \) m s−1. Find its initial acceleration and the distance travelled in the first 10 seconds. [5]
\( a = -e^{-0.2t} \), so initially \( a = -1 \) m s−2. \( v > 0 \) throughout, so distance \( = \int_{0}^{10} 5e^{-0.2t}\,dt = \Big[-25e^{-0.2t}\Big]_{0}^{10} = 25(1 - e^{-2}) = 21.6 \) m.
5. (Calculator.) A particle has velocity \( v = t^{2} - 6t + 8 \) m s−1 for \( 0 \le t \le 5 \). Find the total distance travelled and the final displacement from the starting point. [6]
At rest at \( t = 2 \) and \( t = 4 \). \( \int_{0}^{2} v\,dt = \tfrac{20}{3} \), \( \int_{2}^{4} v\,dt = -\tfrac{4}{3} \), \( \int_{4}^{5} v\,dt = \tfrac{4}{3} \). Distance \( = \tfrac{20}{3} + \tfrac{4}{3} + \tfrac{4}{3} = \tfrac{28}{3} = 9.33 \) m. Displacement \( = \tfrac{20}{3} - \tfrac{4}{3} + \tfrac{4}{3} = \tfrac{20}{3} = 6.67 \) m.
6. (No calculator.) For \( v = 3(t - 1)(t - 3) \), \( 0 \le t \le 4 \), describe the velocity–time and speed–time graphs, giving the key values. [4]
Velocity–time: a parabola, \( v = 9 \) at \( t = 0 \), zero at \( t = 1 \) and \( t = 3 \), minimum \( -3 \) at \( t = 2 \), back to 9 at \( t = 4 \). Speed–time: the same curve with the part between \( t = 1 \) and \( t = 3 \) reflected above the axis, so it has a maximum of 3 at \( t = 2 \) and sharp corners at \( t = 1 \) and \( t = 3 \).
The velocity-time graph v = 3(t minus 1)(t minus 3) for t from 0 to 4: it starts at 9, crosses zero at t = 1 and t = 3, reaches minus 3 at t = 2 and returns to 9 at t = 4. The speed-time graph, drawn dashed, reflects the part between t = 1 and t = 3 above the axis, giving a maximum of 3.
Velocity (solid) and speed (dashed): identical except between \( t = 1 \) and \( t = 3 \), where speed is the reflection.

🔗Go deeper — other people’s work

These are external resources, not mine. If one stops working, tell me and everything above it on this page still stands.

  • Desmos — plot \(s\), \(v\) and \(a\) on the same axes to see how the zeros of one line up with the turning points of another
  • PhET “The Moving Man” — drag a figure and watch position, velocity and acceleration graphs draw themselves