# Part (a) – Time and Distance to Overtake the Bus
| Step | Derivation/Formula | Reasoning |
|---|---|---|
| 1 | x_{\text{bus}}(t) = 0.5 \cdot a \cdot t^2 | For the bus, the starting position is x = 0 , and it accelerates from rest (initial velocity u = 0 ), thus its position as a function of time by integrating acceleration a is x = \frac{1}{2}at^2 . |
| 2 | x_{\text{student}}(t) = 40 + v \cdot t | For the student, the initial position is 40 meters behind the bus, and she runs at a constant speed v = 5.0 \, \text{m/s} , hence moving a distance v \cdot t + 40 in time t . |
| 3 | 0.5 \cdot 0.170 \cdot t^2 = 40 + 5.0 \cdot t | Setting the position equations of the bus and the student equal to each other, as they must meet at the same point when the student catches the bus. |
| 4 | 0.085 \cdot t^2 – 5.0 \cdot t – 40 = 0 | Simplifying the equation yields a quadratic form At^2 + Bt + C = 0 . |
| 5 | t = \frac{-B \pm \sqrt{B^2 – 4AC}}{2A} | Solve for t using the quadratic formula where A = 0.085 , B = -5.0 , C = -40 . |
| 6 | t \approx 28.0 \, \text{s} | Calculate the discriminant and solve for t . Positive root represents the time when the student first catches up to the bus. |
| 7 | x_{\text{student}} = 40 + 5.0 \times 28.0 = 180.0 \, \text{m} | Substitute t back into the student’s position equation to find the distance she needs to run. |
| 8 | Total Time: 28.0 s, Distance: 180.0 m | These values reflect the time and distance required for the student to run from her starting point to catch the bus. |
# Part (b) – Bus Speed When Overtaken
| Step | Derivation/Formula | Reasoning |
|---|---|---|
| 1 | v_{\text{bus}} = a \cdot t | The bus starts from rest and accelerates at a = 0.170 \, \text{m/s}^2 . The velocity at any time t can be found by multiplying acceleration and time. |
| 2 | v_{\text{bus}} = 0.170 \cdot 28.0 | Plug in the time t = 28.0 seconds into the velocity equation for the bus. |
| 3 | v_{\text{bus}} = 4.76 \, \text{m/s} | Calculate the final velocity of the bus. |
| 4 | Bus Speed: 4.76 m/s | This is the speed of the bus at the moment when the student overtakes it. |
# Part (c), (d), (e), (f) will follow in the next messages, keeping each analysis clear and separated.
Phy can also check your working. Just snap a picture!
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| Kinematics | Forces |
|---|---|
| \Delta x = v_i t + \frac{1}{2} at^2 | F = ma |
| v = v_i + at | F_g = \frac{G m_1m_2}{r^2} |
| a = \frac{\Delta v}{\Delta t} | f = \mu N |
| R = \frac{v_i^2 \sin(2\theta)}{g} |
| Circular Motion | Energy |
|---|---|
| F_c = \frac{mv^2}{r} | KE = \frac{1}{2} mv^2 |
| a_c = \frac{v^2}{r} | PE = mgh |
| KE_i + PE_i = KE_f + PE_f |
| Momentum | Torque and Rotations |
|---|---|
| p = m v | \tau = r \cdot F \cdot \sin(\theta) |
| J = \Delta p | I = \sum mr^2 |
| p_i = p_f | L = I \cdot \omega |
| Simple Harmonic Motion |
|---|
| F = -k x |
| T = 2\pi \sqrt{\frac{l}{g}} |
| T = 2\pi \sqrt{\frac{m}{k}} |
| Constant | Description |
|---|---|
| g | Acceleration due to gravity, typically 9.8 , \text{m/s}^2 on Earth’s surface |
| G | Universal Gravitational Constant, 6.674 \times 10^{-11} , \text{N} \cdot \text{m}^2/\text{kg}^2 |
| \mu_k and \mu_s | Coefficients of kinetic (\mu_k) and static (\mu_s) friction, dimensionless. Static friction (\mu_s) is usually greater than kinetic friction (\mu_k) as it resists the start of motion. |
| k | Spring constant, in \text{N/m} |
| M_E = 5.972 \times 10^{24} , \text{kg} | Mass of the Earth |
| M_M = 7.348 \times 10^{22} , \text{kg} | Mass of the Moon |
| M_M = 1.989 \times 10^{30} , \text{kg} | Mass of the Sun |
| Variable | SI Unit |
|---|---|
| s (Displacement) | \text{meters (m)} |
| v (Velocity) | \text{meters per second (m/s)} |
| a (Acceleration) | \text{meters per second squared (m/s}^2\text{)} |
| t (Time) | \text{seconds (s)} |
| m (Mass) | \text{kilograms (kg)} |
| Variable | Derived SI Unit |
|---|---|
| F (Force) | \text{newtons (N)} |
| E, PE, KE (Energy, Potential Energy, Kinetic Energy) | \text{joules (J)} |
| P (Power) | \text{watts (W)} |
| p (Momentum) | \text{kilogram meters per second (kgm/s)} |
| \omega (Angular Velocity) | \text{radians per second (rad/s)} |
| \tau (Torque) | \text{newton meters (Nm)} |
| I (Moment of Inertia) | \text{kilogram meter squared (kgm}^2\text{)} |
| f (Frequency) | \text{hertz (Hz)} |
General Metric Conversion Chart
Example of using unit analysis: Convert 5 kilometers to millimeters.
Start with the given measurement: \text{5 km}
Use the conversion factors for kilometers to meters and meters to millimeters: \text{5 km} \times \frac{10^3 \, \text{m}}{1 \, \text{km}} \times \frac{10^3 \, \text{mm}}{1 \, \text{m}}
Perform the multiplication: \text{5 km} \times \frac{10^3 \, \text{m}}{1 \, \text{km}} \times \frac{10^3 \, \text{mm}}{1 \, \text{m}} = 5 \times 10^3 \times 10^3 \, \text{mm}
Simplify to get the final answer: \boxed{5 \times 10^6 \, \text{mm}}
Prefix | Symbol | Power of Ten | Equivalent |
|---|---|---|---|
Pico- | p | 10^{-12} | |
Nano- | n | 10^{-9} | |
Micro- | µ | 10^{-6} | |
Milli- | m | 10^{-3} | |
Centi- | c | 10^{-2} | |
Deci- | d | 10^{-1} | |
(Base unit) | – | 10^{0} | |
Deca- or Deka- | da | 10^{1} | |
Hecto- | h | 10^{2} | |
Kilo- | k | 10^{3} | |
Mega- | M | 10^{6} | |
Giga- | G | 10^{9} | |
Tera- | T | 10^{12} |
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