Physics - Kinematics

1. Scalars and Vectors

Vector Operations

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

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

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

2. Distance and Displacement

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

3. Speed and Velocity

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

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

Displacement-Time Graphs

Velocity-Time Graphs

4. Acceleration

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

5. Equations of Motion (Uniform Acceleration)

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)

Problem-Solving Strategy

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)

6. Acceleration Due to Gravity

Free Fall

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

Key Principles

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

Experiments to Determine $g$

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 $g$

7. Projectile Motion

Principles

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.

Equations

Horizontal: $x = v_x \cdot t$

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

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

Key Results

• 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 $60°$)

Trajectory

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).

8. Stopping Distance

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

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

Thinking 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)

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

Braking Distance

• 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

Factors Affecting Stopping Distance

Typical Values

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.

9. Key Equations Reference

Worked Examples

Example 1: Solving a Projectile Motion Problem

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: $R = v_x \times t = 13.0 \times 1.53 = 19.9\ \text{m}$.

Example 2: Stopping Distance Calculation

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 $= 12 + 33.3 = 45.3\ \text{m}$.

Common Pitfalls

• 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.

Summary

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.