| Quantity | Symbol | SI Unit | Key Relation |
|---|---|---|---|
| Displacement | \(\Delta x\) | m | Vector; signed |
| Velocity | v | m s\(^{-1}\) | \(v=\dfrac{dx}{dt}\) |
| Acceleration | a | m s\(^{-2}\) | \(a=\dfrac{dv}{dt}\) |
| Force | F | N (kg m s\(^{-2}\)) | \(F=ma\) |
| Work | W | J (N m) | \(W=F\,s\cos\theta\) |
| Power | P | W (J s\(^{-1}\)) | \(P=\dfrac{W}{t}\) |
| Energy | E | J | \(Ek=\tfrac12mv^2,\;Eg=mgh\) |
| Syllabus Block | Topic(s) | Relevance to Displacement‑from‑\(v\!-\!t\) |
|---|---|---|
| 2 – Kinematics | Velocity‑time graphs, uniform & non‑uniform acceleration, projectile motion | Core content – area under the curve gives \(\Delta x\); also used for horizontal range of a projectile. |
| 3 – Dynamics | Newton’s laws, impulse–momentum theorem | Impulse \(\displaystyle \int F\,dt\) mirrors the area method; constant net force → straight‑line \(v\!-\!t\) graph. |
| 4 – Forces, Density & Pressure | Free‑fall, hydrostatic pressure | Free‑fall produces a linear \(v\!-\!t\) graph (slope \(-g\)). |
| 5 – Work, Energy & Power | Work‑energy theorem, power | Area under a force–distance graph = work; analogous to area under a velocity–time graph = displacement. |
| 6 – Deformation of Solids | Hooke’s law, elastic potential energy | Understanding signed areas helps with energy stored in a spring (\(\tfrac12kx^2\)). |
| 7‑8 – Waves & Superposition | Wave speed, pulse motion | Particle displacement on a string follows the same area principle when integrating velocity. |
| 9‑10 – Electricity & DC Circuits | Current, charge, capacitor charging | Charge transferred = \(\int I\,dt\); voltage‑time area gives energy stored in a capacitor. |
| 11 – Particle Physics | Radioactive decay | Area under activity‑time curve gives total decays, reinforcing integration concepts. |
The instantaneous velocity is the time derivative of displacement:
\[
v(t)=\frac{dx}{dt}\qquad\Longrightarrow\qquad
\Delta x=\int{t1}^{t_2}v(t)\,dt
\]
Graphically, \(\displaystyle \int v\,dt\) is the signed area between the curve and the time axis.
| Graph Shape | Motion Description | Displacement Formula (signed) |
|---|---|---|
| Horizontal line | Constant velocity \(v\) | \(\Delta x=v\,(t2-t1)\) |
| Straight line through the origin | Uniform acceleration from rest (\(a\)) | \(\Delta x=\tfrac12 a\,(t2^{2}-t1^{2})\) |
| Straight line not through the origin | Uniform acceleration with initial velocity \(v0\) | \(\Delta x=v0(t2-t1)+\tfrac12 a\,(t2^{2}-t1^{2})\) |
| Triangle (entirely above axis) | Acceleration to a maximum speed, then stop (no reversal) | \(\Delta x=\tfrac12 v{\max}\,t{\text{up}}\) |
| Triangle crossing the axis | Accelerate, reverse, then decelerate | \(\Delta x=\tfrac12 v{\max}t{\text{up}}-\tfrac12|v{\min}|t{\text{down}}\) |
| Trapezium (single sign) | Two different constant velocities or linear change that stays on one side of the axis | \(\Delta x=\tfrac12 (v1+v2)(t2-t1)\) |
| Mixed‑sign trapezium | Piece‑wise linear with a reversal | Sum of two signed trapezia (above + below). |
| Curved line (non‑linear) | Variable acceleration | Integrate analytically or approximate with many thin rectangles. |
Problem: A car starts from rest, accelerates uniformly to \(20\ \text{m s}^{-1}\) in \(5\ \text{s}\), then travels at this speed for a further \(10\ \text{s}\). Find the total displacement.
\[
A_{\text{tri}}=\tfrac12\times5\ \text{s}\times20\ \text{m s}^{-1}=50\ \text{m}.
\]
\[
A_{\text{rect}}=10\ \text{s}\times20\ \text{m s}^{-1}=200\ \text{m}.
\]
\[
\Delta x=50\ \text{m}+200\ \text{m}=250\ \text{m}.
\]
Trapezium → \(\Delta x=\tfrac12(5+15)\times4=40\ \text{m}\).
Integration: \(\displaystyle \Delta x=\int0^5(4t-6)dt=[2t^{2}-6t]0^5=20\ \text{m}\).
Zero‑velocity at \(t=1.5\ \text{s}\) → direction reversal after this instant.
Rectangle: \(8\times3=24\ \text{m}\).
Positive triangle: \(\tfrac12\times3\times8=12\ \text{m}\).
Negative triangle: \(-\tfrac12\times3\times4=-6\ \text{m}\).
Net \(\Delta x = 24+12-6 = 30\ \text{m}\).
The “area = change of quantity” idea recurs throughout A‑Level physics. Three concrete examples are given below.
Impulse \(\displaystyle J=\int{t1}^{t_2}F(t)\,dt\) is the signed area under a force‑time graph.
Worked example: A 0.5 kg ball experiences a constant force of 10 N for 0.2 s.
\(J = 10\ \text{N}\times0.2\ \text{s}=2\ \text{N·s}\).
Since \(J=\Delta p\), the ball’s momentum changes by \(2\ \text{kg·m s}^{-1}\) and its speed increases by \(\Delta v = J/m = 4\ \text{m s}^{-1}\).
Charge transferred \(Q=\displaystyle\int I(t)\,dt\) equals the area under a current‑time graph.
Worked example: A circuit carries a current that linearly rises from 0 A to 3 A in 5 s.
Area (triangle) = \(\tfrac12\times5\ \text{s}\times3\ \text{A}=7.5\ \text{C}\).
Thus 7.5 C of charge pass through the circuit.
Energy stored while a capacitor charges: \(\displaystyle W=\int V(t)I(t)\,dt\).
If a constant voltage of 12 V is applied to a resistor‑capacitor circuit drawing a current that decays as \(I=I_0e^{-t/RC}\), the energy is the area under the \(V\!-\!I\) curve.
For a simple case with constant \(V\) and average current \(\bar I\) over time \(\Delta t\):
\(W = V\bar I\Delta t\).
Example: \(V=12\ \text{V}\), \(\bar I=0.5\ \text{A}\), \(\Delta t=4\ \text{s}\) → \(W=12\times0.5\times4=24\ \text{J}\).
\[
\delta(\Delta x)=\sqrt{(\delta v\,\Delta t)^2+(v\,\delta t)^2}
\]
where \(\delta v\) and \(\delta t\) are the instrument uncertainties.
| Mistake | Why it happens | How to prevent it |
|---|---|---|
| Sign error | Forgetting that areas below the axis are negative. | Mark zero‑velocity points first; colour‑code above (green) and below (red) the axis. |
| Unit mismatch | Mixing km h\(^{-1}\) with seconds. | Convert all velocities to m s\(^{-1}\) and times to seconds before calculating area. |
| Treating curved graphs as straight | Skipping the integration step. | Divide the curve into many thin rectangles (Δt → 0) or use the analytical expression if given. |
| Confusing displacement with distance | Adding absolute values of all areas. | Remember: displacement = signed sum; total distance = sum of absolute areas. |
| Ignoring uncertainties | Only reporting a single value. | Always propagate the uncertainties from the measured quantities and quote \(\pm\) values. |
The signed area under a velocity–time graph provides a quick, visual route to displacement. Mastery of this technique links directly to many other parts of the Cambridge 9702 syllabus, from Newton’s laws to energy concepts, and re‑appears in A‑Level topics such as impulse, charge transfer and capacitor energy. By following the step‑by‑step procedure, applying the correct sign convention, and accounting for uncertainties, students can solve a wide range of exam questions efficiently and accurately.
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