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# 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.

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##### Made By Nerd-Notes.com
KinematicsForces
\Delta x = v_i t + \frac{1}{2} at^2F = ma
v = v_i + atF_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 MotionEnergy
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
MomentumTorque and Rotations
p = m v\tau = r \cdot F \cdot \sin(\theta)
J = \Delta pI = \sum mr^2
p_i = p_fL = I \cdot \omega
Simple Harmonic Motion
F = -k x
T = 2\pi \sqrt{\frac{l}{g}}
T = 2\pi \sqrt{\frac{m}{k}}
ConstantDescription
gAcceleration due to gravity, typically 9.8 , \text{m/s}^2 on Earth’s surface
GUniversal Gravitational Constant, 6.674 \times 10^{-11} , \text{N} \cdot \text{m}^2/\text{kg}^2
\mu_k and \mu_sCoefficients 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.
kSpring 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
VariableSI 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)}
VariableDerived 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)}
\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.

1. Start with the given measurement: \text{5 km}

2. 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}}

3. 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}

4. 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}

1. Some answers may be slightly off by 1% depending on rounding, etc.
2. Answers will use different values of gravity. Some answers use 9.81 m/s2, and other 10 m/s2 for calculations.
3. Variables are sometimes written differently from class to class. For example, sometime initial velocity v_i is written as u ; sometimes \Delta x is written as s .
4. Bookmark questions that you can’t solve so you can come back to them later.
5. Always get help if you can’t figure out a problem. The sooner you can get it cleared up the better chances of you not getting it wrong on a test!

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