Step  Derivation/Formula  Reasoning 

1  \text{Analyzing the Air Resistance Effect}  The presence of air resistance as a significant factor means that the velocity of the beach ball will increase over time but at a decreasing rate. This is because air resistance (drag) opposes the motion of the ball and increases with velocity. 
2  \text{Behavior of Velocity Under Air Resistance}  Typically, with air resistance, the object will eventually reach a terminal velocity where the gravitational force (downwards) is balanced by the drag force (upwards), and acceleration ceases, leading to constant velocity. 
3  \text{Analyzing Each Option} 

4  \text{Conclusion}  (D) best represents the situation of a beach ball dropping with air resistance, as it shows an initial increase in velocity that begins to level off, suggesting the onset of reaching terminal velocity. 
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At time t = 0, a cart is at x = 10 m and has a velocity of 3 m/s in the –x direction. The cart has a constant acceleration in the +x direction with magnitude 3 m/s^{2} < a < 6 m/s^{2}?. Which of the following gives the possible range of the position of the cart at t = 1 s?
An object is thrown upward at 65 m/s from the top of a 800 m tall building.
A car travels at 20 m/s for 5 minutes and then travels another 2 km at 40 m/s. What is the total
distance traveled and time of travel for the car?
An object is projected vertically upward from ground level. It rises to a maximum height H . If air resistance is negligible, which of the following must be true for the object when it is at a height H/2 ?
The graph above shows velocity v versus time t for an object in linear motion. Which of the following is a possible graph of position x versus time t for this object?
D
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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}  0.000000000001 
Nano  n  10^{9}  0.000000001 
Micro  µ  10^{6}  0.000001 
Milli  m  10^{3}  0.001 
Centi  c  10^{2}  0.01 
Deci  d  10^{1}  0.1 
(Base unit)  –  10^{0}  1 
Deca or Deka  da  10^{1}  10 
Hecto  h  10^{2}  100 
Kilo  k  10^{3}  1,000 
Mega  M  10^{6}  1,000,000 
Giga  G  10^{9}  1,000,000,000 
Tera  T  10^{12}  1,000,000,000,000 
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