Physics - Kinematics | DSE - Wyatt's Notes

• Adding vectors: use tip-to-tail method or resolve into components
• Resolving a vector: split into horizontal ($x$) and vertical ($y$) components

$F_x = F\cos\theta \quad \text{and} \quad F_y = F\sin\theta$

$F = \sqrt{F_x^2 + F_y^2}$

• Distance is always $\geq$ displacement
• Displacement can be zero if the object returns to its starting point

$v = \frac{\Delta s}{\Delta t}$

Average speed: total distance / total time Instantaneous speed: speed at a specific moment (gradient of distance-time graph at that point)

Acceleration is the rate of change of velocity:

$a = \frac{\Delta v}{\Delta t} = \frac{v - u}{t}$

• Units: $\mathrm{m\,s^{-2}}$
• A positive acceleration means the object is speeding up in the direction of motion
• A negative acceleration (deceleration) means the object is slowing down

For motion with constant acceleration (also called the suvat equations):

Where:

• $s$ = displacement (m)
• $u$ = initial velocity ($\mathrm{m\,s^{-1}}$)
• $v$ = final velocity ($\mathrm{m\,s^{-1}}$)
• $a$ = acceleration ($\mathrm{m\,s^{-2}}$)
• $t$ = time (s)

1. List the known quantities ($s, u, v, a, t$) and identify what to find
2. Choose the suvat equation with exactly those variables (one unknown)
3. Solve the equation
4. Check: does the answer make sense? (e.g. negative displacement, zero final velocity)

All objects in free fall (only gravity acting, no air resistance) accelerate at the same rate: $g \approx 9.81\ \mathrm{m\,s^{-2}}$

1. Near Earth”s surface, $g$ is approximately constant at $9.81\ \mathrm{m\,s^{-2}}$
2. $g$ does not depend on the mass of the falling object (Galileo’s principle)
3. In a vacuum, a feather and a bowling ball fall at the same rate

Method 1: Electromagnet and trapdoor

• Electromagnet holds a steel ball; switch releases it
• Timer starts when ball released; stops when it hits trapdoor
• Repeat for various heights $h$

$h = \frac{1}{2}gt^2 \implies g = \frac{2h}{t^2}$

Plot $h$ vs $t^2$; gradient = $\frac{g}{2}$

Method 2: Light gates

• Measure time for card to pass through two light gates at known separation
• Gives velocity at each gate; use $v^2 = u^2 + 2as$ to find $g$

A projectile moves under the influence of gravity only (air resistance is neglected in DSE).

Key idea: horizontal and vertical motion are independent of each other.

Horizontal: $x = v_x \cdot t$

where $v_x = v_0\cos\theta$ (constant throughout)

Vertical: $v_y = u_y + at = v_0\sin\theta - gt$ $y = u_y t + \frac{1}{2}at^2 = v_0\sin\theta\cdot t - \frac{1}{2}gt^2$

• Time of flight: set $y = 0$ and solve for $t$; or use $t = \frac{2v_0\sin\theta}{g}$
• Maximum height: occurs when $v_y = 0$; $H = \frac{(v_0\sin\theta)^2}{2g}$
• Range: $R = \frac{v_0^2\sin 2\theta}{g}$
• Maximum range: at $\theta = 45°$ (complementary angles give equal ranges: $30°$ and $60°$)

The path of a projectile is a parabola. The vertical velocity is zero at the highest point, but horizontal velocity is never zero (in ideal conditions).

The stopping distance is the total distance a vehicle travels from the moment the driver sees a hazard to the moment the vehicle stops:

$\text{Stopping distance} = \text{Thinking distance} + \text{Braking distance}$

• Distance travelled during the driver’s reaction time before brakes are applied
• Depends on: speed (directly proportional), reaction time (affected by alcohol, drugs, fatigue, mobile phone use)

$\text{Thinking distance} = \text{speed} \times \text{reaction time}$

• Distance travelled after brakes are applied until the vehicle stops
• Depends on: speed (proportional to $v^2$), road conditions (wet/icy), tyre condition, vehicle mass, brake efficiency

At $30\ \mathrm{mph}$: total $\approx 23\ \mathrm{m}$ (9 m thinking + 14 m braking) At $70\ \mathrm{mph}$: total $\approx 96\ \mathrm{m}$ (21 m thinking + 75 m braking)

Note how braking distance increases much more than thinking distance as speed increases.

Problem: A ball is thrown from ground level with initial velocity $15\ \mathrm{m\,s^{-1}}$ at an angle of $30°$ above the horizontal. Find the maximum height and the horizontal range. Solution: Resolve: $v_x = 15\cos 30° = 13.0\ \mathrm{m\,s^{-1}}$, $v_y = 15\sin 30° = 7.5\ \mathrm{m\,s^{-1}}$. Maximum height: $H = \dfrac{v_y^2}{2g} = \dfrac{7.5^2}{2 \times 9.81} = 2.87\ \text{m}$. Time of flight: $t = \dfrac{2v_y}{g} = \dfrac{2 \times 7.5}{9.81} = 1.53\ \text{s}$. Range: $R = v_x \times t = 13.0 \times 1.53 = 19.9\ \text{m}$.

Problem: A car travelling at $20\ \mathrm{m\,s^{-1}}$ (approximately 45 mph) has a reaction time of 0.6 s. Braking deceleration is $6.0\ \mathrm{m\,s^{-2}}$. Calculate the total stopping distance. Solution: Thinking distance $= v \times t_{\text{reaction}} = 20 \times 0.6 = 12\ \text{m}$. For braking: $v^2 = u^2 + 2as$ with $v = 0$, $u = 20$, $a = -6.0$. $0 = 400 + 2(-6.0)s \implies s = 33.3\ \text{m}$. Total stopping distance $= 12 + 33.3 = 45.3\ \text{m}$.

• Using distance instead of displacement in suvat equations: The suvat equations apply to displacement, not total distance travelled. For projectile motion, use the vertical component for height calculations.
• Forgetting that horizontal velocity is constant: In projectile motion, there is no horizontal acceleration (air resistance is neglected in DSE). Do not apply $v = u + at$ to the horizontal component.
• Confusing speed and velocity: Speed is a scalar; velocity is a vector. A ball thrown vertically upward has constant acceleration (downward) even at the highest point where its velocity is momentarily zero.

flowchart TD
    A[Kinematics] --> B[Key Concepts]
    A --> C[Core Principles]
    A --> D[Practical Applications]
    B --> E[Fundamental definitions]
    C --> F[Design patterns]
    D --> G[Real-world usage]

Kinematics covers the distinction between scalars and vectors, the equations of uniformly accelerated motion (suvat), projectile motion (independent horizontal and vertical components), stopping distance (thinking and braking components), and graphical interpretations of motion (displacement-time and velocity-time graphs). The acceleration due to gravity ($g \approx 9.81\ \mathrm{m\,s^{-2}}$) is constant for free fall near Earth’s surface.

Physics reveals that nature follows mathematical laws at every scale. Matter is made of atoms, forces arise from field interactions, and energy is conserved in every transformation. The power of physics lies in its predictive ability - from calculating projectile trajectories to designing particle accelerators. Understanding these principles helps us technology, predict natural phenomena, and appreciate the universe’s underlying order.

• Mechanics
• Waves
• Electricity
• Fields