These notes cover the Cambridge AS & A‑Level Physics (9702) learning outcomes for Topic 2 – Kinematics. They explain the concepts, give the four kinematic equations, show how to use graphs, apply the equations to free‑fall, and provide worked examples, common pitfalls and practice questions.
| Quantity | Symbol | Nature | Unit |
|---|---|---|---|
| Displacement | s | Vector (sign shows direction) | m |
| Distance travelled | – | Scalar (always positive) | m |
| Velocity | v | Vector | m s⁻¹ |
| Speed | – | Scalar (|v|) | m s⁻¹ |
| Acceleration | a | Vector | m s⁻² |
| Time | t | Scalar | s |
Standard syllabus symbols:
A car moves at a constant speed of 15 m s⁻¹ for 4 s. The s‑t graph is a straight line with gradient 15 m s⁻¹. The displacement after 4 s is
\$s = v t = 15 \times 4 = 60\ \text{m}\$
The area under the v‑t graph (a rectangle) gives the same result, confirming the gradient‑velocity link.
Uniform acceleration means a is constant, so
\$a = \frac{dv}{dt} = \frac{v-u}{t}\quad\Longrightarrow\quad v = u + a t \tag{1}\$
Displacement is the integral of velocity:
\$s = \int v\,dt = \int (u + a t)\,dt = u t + \frac12 a t^{2} \tag{2}\$
Eliminate t between (1) and (2). From (1), t = (v-u)/a. Substituting into (2) gives
\[
s = u\frac{v-u}{a} + \frac12 a\left(\frac{v-u}{a}\right)^{2}
= \frac{v^{2}-u^{2}}{2a}
\;\Longrightarrow\; v^{2}=u^{2}+2as \tag{3}
\]
Because acceleration is constant, the average velocity is the arithmetic mean of the initial and final velocities:
\$\bar v = \frac{u+v}{2}\quad\Longrightarrow\quad s = \bar v\,t = \frac12 (u+v) t \tag{4}\$
Equations (1)–(4) are the four core kinematic formulas.
| Equation | Variables present | Typical use |
|---|---|---|
| \$v = u + a t\$ | u, a, t | Find final speed or time when acceleration is known. |
| \$s = u t + \tfrac12 a t^{2}\$ | u, a, t | Displacement after a known time. |
| \$v^{2} = u^{2} + 2 a s\$ | u, a, s | Relate speeds and displacement without needing time. |
| \$s = \tfrac12 (u+v) t\$ | u, v, t | Use when both initial and final speeds are known. |
Replace a by ±g (with g = 9.81 m s⁻²) according to the chosen positive direction:
“+” is used when downwards is taken as positive; “–” when upwards is positive.
A stone is released from rest at a height of 20 m above the ground. Take upwards as positive. Find the time to reach the ground and the impact speed.
\$-20 = 0\cdot t + \tfrac12(-9.81) t^{2}\$
\$t^{2}= \frac{2\times20}{9.81}\approx4.08\;\Rightarrow\;t\approx2.02\ \text{s}\$
\$v = 0 + (-9.81)(2.02)\approx-19.8\ \text{m s}^{-1}\$
Speed = 19.8 m s⁻¹ downwards.
A ball is thrown vertically upwards with an initial speed of 15 m s⁻¹. Take upwards as positive.
\$0 = 15^{2}+2(-9.81)s\;\Rightarrow\;s=\frac{15^{2}}{2\times9.81}\approx11.5\ \text{m}\$
\$0 = 15 + (-9.81) t\;\Rightarrow\;t_{\text{up}}=\frac{15}{9.81}\approx1.53\ \text{s}\$
A car accelerates uniformly from 0 to 20 m s⁻¹ in 5 s. From the straight‑line v–t graph determine the displacement.
A cyclist travels at a constant speed of 8 m s⁻¹ for 6 s. Sketch the s–t graph and read the displacement.
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