Equations of Motion – Cambridge IGCSE / A‑Level Physics (9702)
This topic belongs to Topic 2 Kinematics in the Cambridge syllabus. It builds on Newton’s second law, force and mass, and assumes you are comfortable with those ideas.
1. Kinematic Variables, Symbols & Units
| Quantity | Symbol | Definition | SI Unit |
|---|
| Distance | s | Scalar length of the path travelled | metre (m) |
| Displacement | \(\Delta s\) | Vector change in position (straight‑line) | metre (m) |
| Speed | \(|v|\) | Scalar magnitude of velocity | metre per second (m s⁻¹) |
| Velocity | v | Vector rate of change of displacement, \(v = \frac{ds}{dt}\) | metre per second (m s⁻¹) |
| Acceleration | a | Vector rate of change of velocity, \(a = \frac{dv}{dt}\) | metre per second squared (m s⁻²) |
| Time | t | Independent scalar variable | second (s) |
2. Sign‑Convention Checklist
- Choose a positive direction (e.g. rightwards or upwards) before you start.
- All quantities measured in that direction are taken as positive; opposite‑direction quantities are negative.
- Apply the same sign convention consistently to displacement, velocity, acceleration and any forces that appear later.
- Write the chosen convention at the start of every solution – it saves marks in the exam.
3. Graphical Interpretation of Motion
3.1 What the gradients give
| Graph | Gradient = … |
|---|
| Displacement‑time (s‑t) | Velocity (v) |
| Velocity‑time (v‑t) | Acceleration (a) |
| Acceleration‑time (a‑t) | Rate of change of acceleration (rarely needed) |
3.2 What the areas give
| Graph | Area = … |
|---|
| Displacement‑time (s‑t) | – (no direct physical meaning) |
| Velocity‑time (v‑t) | Displacement, \(\Delta s\) |
| Acceleration‑time (a‑t) | Change in velocity, \(\Delta v\) |
3.3 Worked graphical examples
Example A – v‑t graph (displacement from area)
A particle starts from rest and accelerates uniformly at \(2\;\text{m s}^{-2}\) for \(4\;\text{s}\). The v‑t graph is a straight line from \((0,0)\) to \((4,8)\).
- Gradient \(= a = 2\;\text{m s}^{-2}\).
- Area \(= \frac12 \times 4 \times 8 = 16\;\text{m}\) → displacement.
Example B – a‑t graph (change in velocity from area)
An object experiences a constant acceleration of \(-3\;\text{m s}^{-2}\) for \(5\;\text{s}\). The a‑t graph is a horizontal line at \(-3\) from \(t=0\) to \(t=5\).
- Area \(= (-3)\times5 = -15\;\text{m s}^{-1}\) → \(\Delta v = -15\;\text{m s}^{-1}\) (the speed decreases).
4. Constant‑Acceleration Assumption
All three core equations that follow are derived **only** under the condition that the acceleration a is constant throughout the motion. If a varies, the equations cannot be used and calculus (or piece‑wise analysis) is required. This assumption is explicitly stated in the syllabus and is a common source of loss of marks.
5. Derivation of the Three Core Equations (Constant a)
5.1 From the definition of acceleration
\[
a = \frac{dv}{dt}\quad\Longrightarrow\quad dv = a\,dt
\]
Integrating from \(t=0\) (where \(v=u\)) to \(t\):
\[
\int_{u}^{v} dv = a\int_{0}^{t} dt \;\Longrightarrow\; v = u + at \tag{1}
\]
5.2 From the definition of velocity
\[
v = \frac{ds}{dt}\quad\Longrightarrow\quad ds = v\,dt
\]
Substituting (1) for \(v\) and integrating:
\[
\int_{0}^{s} ds = \int_{0}^{t} (u + at)\,dt
\]
\[
s = ut + \frac12 at^{2} \tag{2}
\]
5.3 Eliminating time
From (1), \(t = \dfrac{v-u}{a}\). Insert this into (2):
\[
s = u\left(\frac{v-u}{a}\right) + \frac12 a\left(\frac{v-u}{a}\right)^{2}
= \frac{v^{2} - u^{2}}{2a}
\]
Re‑arranged:
\[
v^{2} = u^{2} + 2as \tag{3}
\]
6. Summary Table of the Three Equations
| Equation | Form | When to use |
|---|
| First | \(v = u + at\) | Known: \(u, a, t\). Find final velocity (or acceleration). |
| Second | \(s = ut + \tfrac12 at^{2}\) | Known: \(u, a, t\). Find displacement (or acceleration). |
| Third | \(v^{2} = u^{2} + 2as\) | Known: \(u, a, s\). Find final velocity (or displacement) when time is not required. |
7. Common Applications
7.1 Free Fall (vertical motion)
- Take downward as positive (or upward – be consistent).
- Acceleration \(a = g = 9.81\;\text{m s}^{-2}\) (downwards).
- When an object is released from rest, \(u = 0\); the three equations reduce to:
- \(v = gt\)
- \(s = \tfrac12 gt^{2}\)
- \(v^{2} = 2gs\)
7.2 Horizontal projectile motion
Motion is analysed in two independent perpendicular directions.
- Horizontal (x‑direction): \(a_{x}=0\) → \(v_{x}=u_{x}= \text{constant}\).
- Vertical (y‑direction): constant acceleration \(a_{y}=g\) (downwards).
- Use the appropriate equation for each axis, then combine:
\[
\text{Range}=v_{x}\,t_{\text{flight}},\qquad
t_{\text{flight}} = \sqrt{\frac{2h}{g}} \;(\text{if released from height }h)
\]
7.3 Inclined plane (frictionless)
Component of gravity acting along the plane:
\[
a = g\sin\theta
\]
Derivation:
- Weight \(mg\) acts vertically downwards.
- Resolve \(mg\) into components parallel (\(mg\sin\theta\)) and perpendicular (\(mg\cos\theta\)) to the plane.
- With no friction, only the parallel component produces acceleration: \(ma = mg\sin\theta \Rightarrow a = g\sin\theta\).
Use the three equations with this \(a\) and measure all distances \(s\) along the plane.
7.4 Uniform circular motion – tangential acceleration
- The three equations apply **only to the tangential direction** when the speed changes at a constant rate.
- Radial (centripetal) acceleration \(a_{c}=v^{2}/r\) is perpendicular to the motion and is **not** described by these equations.
- Typical A‑Level question: a car speeds up uniformly while negotiating a curve – use the equations for the change in speed, then combine with \(a_{c}\) to find net acceleration.
8. Worked Examples
8.1 Car accelerating from rest
From rest to \(30\;\text{m s}^{-1}\) in \(10\;\text{s}\).
- Equation (1): \(30 = 0 + a(10) \Rightarrow a = 3\;\text{m s}^{-2}\).
- Equation (2): \(s = 0\cdot10 + \tfrac12(3)(10)^{2}=150\;\text{m}\).
8.2 Free‑fall from a cliff
A stone is dropped from rest from a height of \(80\;\text{m}\).
- Use \(s = \tfrac12 gt^{2}\) → \(80 = \tfrac12(9.81)t^{2}\) → \(t = 4.04\;\text{s}\).
- Final speed \(v = gt = 9.81\times4.04 = 39.6\;\text{m s}^{-1}\).
8.3 Horizontal projectile
Launch speed \(u_{x}=20\;\text{m s}^{-1}\) from a \(45\;\text{m}\) high cliff.
- Time of flight \(t = \sqrt{2h/g}= \sqrt{2\times45/9.81}=3.03\;\text{s}\).
- Horizontal range \(R = u_{x}t = 20\times3.03 = 60.6\;\text{m}\).
- Final vertical speed \(v_{y}=gt = 9.81\times3.03 = 29.7\;\text{m s}^{-1}\) (downwards).
- Resultant speed \(v = \sqrt{u_{x}^{2}+v_{y}^{2}} = 36.0\;\text{m s}^{-1}\).
8.4 Block down a frictionless incline
Incline \(\theta = 30^{\circ}\), distance along plane \(s = 5\;\text{m}\), start from rest.
- Acceleration \(a = g\sin30^{\circ}=9.81\times0.5=4.905\;\text{m s}^{-2}\).
- Using (3): \(v^{2}=0+2as = 2(4.905)(5)=49.05\) → \(v = 7.0\;\text{m s}^{-1}\).
8.5 Area under an a‑t graph (change in velocity)
A constant acceleration of \(-2\;\text{m s}^{-2}\) acts for \(6\;\text{s}\).
- Area = \((-2)\times6 = -12\;\text{m s}^{-1}\) → \(\Delta v = -12\;\text{m s}^{-1}\).
- If the initial speed was \(15\;\text{m s}^{-1}\), the final speed is \(3\;\text{m s}^{-1}\) in the original direction.
9. Practice Questions
- A stone is dropped from rest from a cliff \(80\;\text{m}\) high. Calculate the time to reach the ground and the impact speed. (g = 9.81 m s⁻²)
- A projectile is launched horizontally with a speed of \(20\;\text{m s}^{-1}\) from a cliff \(45\;\text{m}\) high. Determine the time of flight, horizontal range and final speed.
- A block slides down a frictionless plane inclined at \(30^{\circ}\) from rest. Find its speed after it has travelled \(5\;\text{m}\) along the plane.
- A car travelling at \(25\;\text{m s}^{-1}\) brakes uniformly to a stop in \(8\;\text{s}\). Compute the magnitude of the deceleration and the distance covered during braking.
- Using the v‑t graph below (a straight line from \((0\,\text{s},0\,\text{m s}^{-1})\) to \((6\,\text{s},12\,\text{m s}^{-1})\)):
- Find the acceleration.
- Find the displacement after the full 6 s.
- Find the speed after 4 s.
- An object experiences a constant acceleration of \(-3\;\text{m s}^{-2}\) for \(5\;\text{s}\). Starting from a speed of \(20\;\text{m s}^{-1}\), determine the final speed and the total distance travelled.
10. Looking Ahead
The kinematic equations you have mastered are the foundation for later topics such as:
- Motion in a circle – linking tangential acceleration (use equations) with centripetal acceleration.
- Newton’s laws of motion – using \(F = ma\) to derive the equations for objects under constant net force.
- Gravitational fields – applying the free‑fall results to orbital motion and escape velocity.
When you encounter these topics, remember to return to the sign‑convention checklist and the constant‑acceleration assumption – they remain essential throughout the syllabus.
11. Summary
- The three equations of motion (1)–(3) are valid **only for constant acceleration**.
- Speed \(|v|\) is the scalar magnitude of the vector velocity v.
- Graphical methods:
- Gradient → instantaneous velocity or acceleration.
- Area → displacement (v‑t) or change in velocity (a‑t).
- Choose a consistent sign convention before solving any problem.
- Select the appropriate equation:
- Equation (1) when time is known.
- Equation (2) when displacement is required.
- Equation (3) when time is not needed.
- Keep units consistent, carry the correct number of significant figures, and always state the direction (sign) of vector quantities.