Kinematics

Kinematics

Info: Board Coverage AQA Paper 1 | Edexcel CP1, CP2 | OCR (A) Paper 1 | CIE P2

1. Fundamental Definitions

Kinematics is the mathematical description of motion, without reference to the forces that cause it. We begin with three rigorous definitions.

Definition. The displacement $s$ of a particle is its position vector relative to a chosen Origin. Unlike distance, displacement is a vector quantity.

Definition. The velocity $v$ of a particle is the rate of change of its displacement with Respect to time:

$v = \frac{ds}{dt} = \lim_{\Delta t \to 0}\frac{\Delta s}{\Delta t}$

Velocity is a vector. Its magnitude is the speed.

Definition. The acceleration $a$ of a particle is the rate of change of its velocity with Respect to time:

$a = \frac{dv}{dt} = \frac{d^2s}{dt^2}$

Acceleration is also a vector. The SI unit is m s$^{-2}$.

2. Derivation of the SUVAT Equations

For uniform acceleration (constant $a$), we derive five equations relating the kinematic Variables $s$, $u$, $v$, $a$And $t$.

Equation 1: $v = u + at$

Starting from the definition of acceleration:

$a = \frac{dv}{dt}$

Since $a$ is constant, we integrate with respect to time:

$\int_{u}^{v} dv = \int_{0}^{t} a \, dt' \implies v - u = at \implies \boxed{v = u + at}$

Equation 2: $s = \frac{1}{2}(u + v)t$

From the definition of velocity:

$v = \frac{ds}{dt}$

For constant acceleration, the velocity varies linearly from $u$ to $v$. The average velocity over The interval is $\bar{v} = \frac{u + v}{2}$. Since displacement equals average velocity multiplied By time:

$\boxed{s = \frac{1}{2}(u + v)t}$

Rigorous derivation. Integrating $ds = v \, dt$ and substituting $v = u + at$:

$s = \int_0^t (u + at')\,dt' = ut + \frac{1}{2}at^2$

From Equation 1, $v = u + at$So $at = v - u$Giving:

$s = ut + \frac{1}{2}(v-u)t = \frac{1}{2}(u + v)t \quad \square$

Equation 3: $s = ut + \frac{1}{2}at^2$

This was obtained above during the rigorous derivation of Equation 2:

$\boxed{s = ut + \frac{1}{2}at^2}$

Equation 4: $v^2 = u^2 + 2as$

We eliminate $t$ between Equations 1 and 3. From Equation 1: $t = \frac{v - u}{a}$. Substituting Into Equation 3:

$\boxed{v^2 = u^2 + 2as}$

Equation 5: $s = vt - \frac{1}{2}at^2$

Substitute $u = v - at$ (from Equation 1) into Equation 3:

$s = (v - at)t + \frac{1}{2}at^2 = vt - at^2 + \frac{1}{2}at^2$

$\boxed{s = vt - \frac{1}{2}at^2}$

3. Displacement-Time and Velocity-Time Graphs

• The gradient of a displacement-time graph gives the velocity.
• The gradient of a velocity-time graph gives the acceleration.
• The area under a velocity-time graph gives the displacement.

Explore the simulation above to develop intuition for this topic.

For uniform acceleration, the $v$-$t$ graph is a straight line, and the area is a trapezium.

For non-constant acceleration, the $v$-$t$ graph is curved, but the same three principles apply At every instant:

• The gradient of the $s$-$t$ curve at any point gives the instantaneous velocity at that time.
• The gradient of the $v$-$t$ curve at any point gives the instantaneous acceleration at that time.
• The area under the $v$-$t$ curve between two times gives the displacement over that interval.

When the $v$-$t$ graph is curved, the area cannot be found using the trapezium rule for a single Straight line. Two approaches are available:

1. Counting squares — estimate the area by counting grid squares on the graph paper, treating partial squares by eye.
2. Integration — if $v(t)$ is given algebraically, compute $\int_{t_1}^{t_2} v(t)\,dt$ exactly.

4. Free Fall

A body in free fall moves under the influence of gravity alone (neglecting air resistance). Near The Earth’s surface, all objects experience the same gravitational acceleration:

$g \approx 9.81 \mathrm{ m s}^{-2}$

This was established by Galileo’s experiments and is a consequence of the equivalence principle (mass cancels in $F = ma = mg$).

Info: Info Problems. Always use the value specified in the question.

5. Projectile Motion

Assumptions

We assume:

1. The only force acting is gravity (no air resistance).
2. $g$ is constant (valid for trajectories small compared to Earth’s radius).
3. The horizontal and vertical components of motion are independent.

Independence of Components

This is the central insight. Since gravity acts vertically, it produces no horizontal Acceleration. Therefore:

• Horizontal: $a_x = 0 \implies v_x = \mathrm{const} = v_0\cos\theta$
• Vertical: $a_y = -g \implies$ the motion is uniformly accelerated

Deriving the Parabolic Trajectory

A projectile is launched from the origin with speed $v_0$ at angle $\theta$ above the horizontal.

Horizontal motion (constant velocity):

$x = v_0\cos\theta \cdot t \implies t = \frac{x}{v_0\cos\theta}$

Vertical motion (uniform acceleration):

$y = v_0\sin\theta \cdot t - \frac{1}{2}gt^2$

Substituting $t$ from the horizontal equation:

This is the equation of a parabola — a direct consequence of constant horizontal velocity Combined with constant vertical acceleration.

Maximum Height

At maximum height, the vertical velocity is zero: $v_y = v_0\sin\theta - gt = 0$Giving $t_{\mathrm{peak}} = \frac{v_0\sin\theta}{g}$.

$H = v_0\sin\theta \cdot \frac{v_0\sin\theta}{g} - \frac{1}{2}g\left(\frac{v_0\sin\theta}{g}\right)^2 = \frac{v_0^2\sin^2\theta}{2g}$

$\boxed{H = \frac{v_0^2\sin^2\theta}{2g}}$

Time of Flight

The projectile returns to $y = 0$ when:

$0 = v_0\sin\theta \cdot t - \frac{1}{2}gt^2 \implies t(v_0\sin\theta - \frac{1}{2}gt) = 0$

The non-trivial solution gives:

$\boxed{T = \frac{2v_0\sin\theta}{g}}$

Maximum Range

The range is $R = v_0\cos\theta \cdot T$:

$R = v_0\cos\theta \cdot \frac{2v_0\sin\theta}{g} = \frac{v_0^2 \cdot 2\sin\theta\cos\theta}{g} = \frac{v_0^2\sin 2\theta}{g}$

$\boxed{R = \frac{v_0^2\sin 2\theta}{g}}$

Maximum range occurs when $\sin 2\theta = 1$I.e., $\theta = 45^\circ$:

$R_{\max} = \frac{v_0^2}{g}$

Intuition: Why Parabolas?

A parabola arises because the vertical position depends quadratically on time ($y \propto t^2$), While the horizontal position depends linearly on time ($x \propto t$). Eliminating $t$ between a Linear and quadratic relation always produces a parabola. The shape is not special to gravity — it Is the geometry of combining uniform motion in one direction with uniformly accelerated motion in The perpendicular direction.

Effect of Air Resistance on Projectiles

In reality, air resistance (drag) opposes the velocity of a moving object. The magnitude of the drag Force depends on the object’s speed and shape, and increases with speed.

Effect on projectile trajectory. When drag is included:

• The horizontal velocity is no longer constant — it decreases throughout the flight because drag has a horizontal component opposing the motion.
• On the way up, both gravity and the vertical component of drag act downward, so the vertical deceleration is greater than $g$.
• On the way down, gravity acts downward but drag acts upward (opposing the downward velocity), so the vertical acceleration is less than $g$.
• The trajectory is no longer parabolic. The descent is steeper than the ascent, the range is shorter, and the maximum height is lower compared to the idealised case.

Terminal velocity. For an object falling vertically under gravity with air resistance, the drag Force increases with speed. Eventually, the drag force equals the weight of the object:

$F_{\mathrm{drag}} = mg$

At this point the net force is zero, the acceleration is zero, and the object falls at a constant Speed called the terminal velocity $v_T$:

$\boxed{v_T \mathrm{ is reached when } F_{\mathrm{drag}} = mg \implies a = 0}$

The $v$-$t$ graph for a falling object reaching terminal velocity shows the velocity increasing with A decreasing gradient (decreasing acceleration) until it asymptotically approaches $v_T$.

Info: Board Coverage AQA Paper 1 | Edexcel CP2 | OCR (A) Paper 1 | CIE P2

6. Non-Uniform Acceleration

When acceleration is not constant, the SUVAT equations do not apply. Instead, we use calculus to Relate displacement, velocity, and acceleration.

Info: Board Coverage AQA Paper 1 | Edexcel CP2 | OCR (A) Paper 1 | CIE P1, P2

The Differential Relations

From the definitions:

$\boxed{v = \frac{ds}{dt}}$

$\boxed{a = \frac{dv}{dt} = \frac{d^2s}{dt^2}}$

A third relation follows from the chain rule. Since $a = dv/dt$ and $v = ds/dt$:

$a = \frac{dv}{dt} = \frac{dv}{ds} \cdot \frac{ds}{dt} = v\,\frac{dv}{ds}$

$\boxed{a = v\,\frac{dv}{ds}}$

This form is useful when acceleration is given as a function of displacement $s$ rather than time $t$.

Selecting the Correct Form

Derivation of $v^2 = u^2 + 2as$ by Integration

Starting from $a = v\,dv/ds$ and assuming constant $a$:

$\int_{u}^{v} v'\,dv' = \int_{0}^{s} a\,ds'$

$\left[\frac{v'^2}{2}\right]_{u}^{v} = as$

$\frac{v^2}{2} - \frac{u^2}{2} = as$

$\boxed{v^2 = u^2 + 2as} \quad \square$

This derivation relies only on the chain-rule identity $a = v\,dv/ds$.

7. Relative Velocity in Two Dimensions

When two objects move in different directions, their velocities are combined using vector addition. The velocity of A relative to B is:

$\boxed{\mathbf{v}_{AB} = \mathbf{v}_A - \mathbf{v}_B}$

This is the velocity A appears to have when observed from B’s frame of reference.

Relative Velocity and Closest Approach

Two objects A and B are on a collision course if their relative velocity $\mathbf{v}_{AB}$ is Directed along the line joining them. If not, the closest approach occurs when the position vector From A to B is perpendicular to the relative velocity: $\mathbf{r}_{BA} \cdot \mathbf{v}_{BA} = 0$.

Problem Set

Common Pitfalls

1. Forgetting to include units in final answers, especially when working with derived units like $\text{N}\,\text{kg}^{-1}\,\text{m}^2$.

2. Neglecting air resistance or assuming ideal conditions when the question specifies a real-world scenario.

3. Incorrectly applying $\vec{F} = m\vec{a}$when forces are not collinear. Resolve into components first.

4. Rounding intermediate answers too early, which compounds errors in multi-step calculations.

Common Mistakes

1. Applying SUVAT equations to non-uniform acceleration. The five SUVAT equations ONLY apply when acceleration is CONSTANT. If a question involves a changing force (e.g., a spring, air resistance), acceleration is not constant and SUVAT cannot be used. Use calculus (integration) or energy methods instead.

2. Confusing displacement with distance. Displacement is a VECTOR (includes direction) and can be negative. Distance is a SCALAR and is always positive. In vertical motion, upward displacement is positive and downward is negative. A ball thrown up and caught at the same point has zero total displacement but non-zero distance travelled.

3. Taking g as positive in both directions. When using SUVAT with vertical motion, be consistent with sign convention. If upward is positive, then g = -9.81 m/s² (acceleration is downward). Substituting g = +9.81 with upward-positive convention will give the wrong answer for velocity and time.

4. Forgetting that air resistance changes the motion. In many projectile questions, air resistance is neglected. But if a question mentions air resistance, remember: the horizontal component of velocity DECREASES (it is no longer constant), the time of flight DECREASES, and the trajectory is no longer a perfect parabola.

5. Not resolving vectors into components before adding. Velocity, displacement, and acceleration are vectors. You cannot add magnitudes directly unless the vectors are in the same direction. Always resolve into horizontal and vertical components first.

Summary

The key principles covered in this topic are linked in the sub-pages above. Focus on understanding the definitions, applying the formulas or frameworks, and evaluating strengths and limitations of each approach.

Worked Examples

Worked examples demonstrating the application of key concepts are covered in the detailed sub-pages linked above.

Cross-References

• Dynamics: Builds on kinematic descriptions by explaining the forces that cause motion and acceleration.
• Work, Energy and Power: Connects velocity and displacement to energy concepts, showing how motion relates to energy transfer.
• Projectile Motion: Applies kinematic equations to objects launched with both horizontal and vertical components of motion.
• Graphical Analysis: Explores how displacement-time and velocity-time graphs visually represent kinematic relationships.

Intuition

Think of kinematics as the “grammar” of motion. Just as grammar describes how sentences are structured without worrying about their meaning, kinematics describes how objects move without worrying about why they move. The SUVAT equations are your toolkit: each equation links four of the five kinematic variables, so you always pick the one that connects what you know to what you need.

The key insight is that horizontal and vertical motions are independent. A ball thrown horizontally from a cliff has two separate stories happening simultaneously: horizontally, it travels at constant velocity (no horizontal force); vertically, it accelerates downward under gravity. The time of flight is determined entirely by the vertical story. This independence is why projectile paths are parabolas — the horizontal displacement grows linearly with time while the vertical displacement grows quadratically.