AP Physics

Unit 1 - Vectors and Kinematics

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Step Derivation/Formula Reasoning
1 \( v_i = 0 \, \text{m/s} \) The car starts from rest, so the initial velocity is zero.
2 \( a_1 = 4.3 \, \text{m/s}^2 \) The car accelerates with an acceleration of \(4.3 \, \text{m/s}^2\).
3 \( t_1 = 6.8 \, \text{s} \) The car accelerates for a time of \(6.8 \, \text{s}\).
4 \( v_x = v_i + a_1 t_1 \) Use the formula for final velocity after a time interval, where \( v_i \) is the initial velocity, \( a_1 \) is the acceleration, and \( t_1 \) is the time.
5 \( v_x = 0 + (4.3 \, \text{m/s}^2)(6.8 \, \text{s}) \) Plug in the values for the initial velocity, acceleration, and time.
6 \( v_x = 29.24 \, \text{m/s} \) Calculate the final velocity after the acceleration phase.
7 \( d_1 = v_i t_1 + \frac{1}{2} a_1 t_1^2 \) Use the formula for distance traveled during the acceleration phase, where \( v_i \) is the initial velocity, \( a_1 \) is the acceleration, and \( t_1 \) is the time.
8 \( d_1 = 0 \cdot 6.8 + \frac{1}{2} (4.3) (6.8)^2 \) Set initial velocity term to zero and plug in the values for acceleration and time.
9 \( d_1 = 99.292 \, \text{m} \) Calculate the distance traveled during the acceleration phase.
10 \( a_2 = -5.1 \, \text{m/s}^2 \) The car decelerates with an acceleration of \( -5.1 \, \text{m/s}^2 \) (since the car is slowing down).
11 \( v_{x2} = 0 \, \text{m/s} \) Final velocity when the car comes to rest again.
12 \( v_x^2 = v_i^2 + 2 a_2 \Delta x_2 \) Use the kinematic equation relating initial velocity, final velocity, acceleration, and distance traveled.
13 \( 0 = (29.24)^2 + 2(-5.1) \Delta x_2 \) Plug in the values for initial velocity during deceleration, final velocity, and acceleration.
14 \( 0 = 855.2976 – 10.2 \Delta x_2 \) Calculate the term \( (29.24)^2 \) and simplify the equation.
15 \( \Delta x_2 = \frac{855.2976}{10.2} \) Rearrange the equation to solve for the distance traveled during deceleration \( \Delta x_2 \).
16 \( \Delta x_2 = 83.95 \, \text{m} \) Calculate the distance traveled during the deceleration phase.
17 \( d_{\text{total}} = d_1 + \Delta x_2 \) Sum the distances traveled during the acceleration phase and deceleration phase to find the total distance.
18 \( d_{\text{total}} = 99.292 + 83.95 \) Plug in the values for the distances calculated in both phases.
19 \( d_{\text{total}} = 183.242 \, \text{m} \) Calculate the total distance traveled by the car and box the final answer.

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\(183.24 \,\text{m}\)

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KinematicsForces
\(\Delta x = v_i t + \frac{1}{2} at^2\)\(F = ma\)
\(v = v_i + at\)\(F_g = \frac{G m_1 m_2}{r^2}\)
\(v^2 = v_i^2 + 2a \Delta x\)\(f = \mu N\)
\(\Delta x = \frac{v_i + v}{2} t\)\(F_s =-kx\)
\(v^2 = v_f^2 \,-\, 2a \Delta x\) 
Circular MotionEnergy
\(F_c = \frac{mv^2}{r}\)\(KE = \frac{1}{2} mv^2\)
\(a_c = \frac{v^2}{r}\)\(PE = mgh\)
\(T = 2\pi \sqrt{\frac{r}{g}}\)\(KE_i + PE_i = KE_f + PE_f\)
 \(W = Fd \cos\theta\)
MomentumTorque and Rotations
\(p = mv\)\(\tau = r \cdot F \cdot \sin(\theta)\)
\(J = \Delta p\)\(I = \sum mr^2\)
\(p_i = p_f\)\(L = I \cdot \omega\)
Simple Harmonic MotionFluids
\(F = -kx\)\(P = \frac{F}{A}\)
\(T = 2\pi \sqrt{\frac{l}{g}}\)\(P_{\text{total}} = P_{\text{atm}} + \rho gh\)
\(T = 2\pi \sqrt{\frac{m}{k}}\)\(Q = Av\)
\(x(t) = A \cos(\omega t + \phi)\)\(F_b = \rho V g\)
\(a = -\omega^2 x\)\(A_1v_1 = A_2v_2\)
ConstantDescription
[katex]g[/katex]Acceleration due to gravity, typically [katex]9.8 , \text{m/s}^2[/katex] on Earth’s surface
[katex]G[/katex]Universal Gravitational Constant, [katex]6.674 \times 10^{-11} , \text{N} \cdot \text{m}^2/\text{kg}^2[/katex]
[katex]\mu_k[/katex] and [katex]\mu_s[/katex]Coefficients of kinetic ([katex]\mu_k[/katex]) and static ([katex]\mu_s[/katex]) friction, dimensionless. Static friction ([katex]\mu_s[/katex]) is usually greater than kinetic friction ([katex]\mu_k[/katex]) as it resists the start of motion.
[katex]k[/katex]Spring constant, in [katex]\text{N/m}[/katex]
[katex] M_E = 5.972 \times 10^{24} , \text{kg} [/katex]Mass of the Earth
[katex] M_M = 7.348 \times 10^{22} , \text{kg} [/katex]Mass of the Moon
[katex] M_M = 1.989 \times 10^{30} , \text{kg} [/katex]Mass of the Sun
VariableSI Unit
[katex]s[/katex] (Displacement)[katex]\text{meters (m)}[/katex]
[katex]v[/katex] (Velocity)[katex]\text{meters per second (m/s)}[/katex]
[katex]a[/katex] (Acceleration)[katex]\text{meters per second squared (m/s}^2\text{)}[/katex]
[katex]t[/katex] (Time)[katex]\text{seconds (s)}[/katex]
[katex]m[/katex] (Mass)[katex]\text{kilograms (kg)}[/katex]
VariableDerived SI Unit
[katex]F[/katex] (Force)[katex]\text{newtons (N)}[/katex]
[katex]E[/katex], [katex]PE[/katex], [katex]KE[/katex] (Energy, Potential Energy, Kinetic Energy)[katex]\text{joules (J)}[/katex]
[katex]P[/katex] (Power)[katex]\text{watts (W)}[/katex]
[katex]p[/katex] (Momentum)[katex]\text{kilogram meters per second (kgm/s)}[/katex]
[katex]\omega[/katex] (Angular Velocity)[katex]\text{radians per second (rad/s)}[/katex]
[katex]\tau[/katex] (Torque)[katex]\text{newton meters (Nm)}[/katex]
[katex]I[/katex] (Moment of Inertia)[katex]\text{kilogram meter squared (kgm}^2\text{)}[/katex]
[katex]f[/katex] (Frequency)[katex]\text{hertz (Hz)}[/katex]

General Metric Conversion Chart

Example of using unit analysis: Convert 5 kilometers to millimeters. 

  1. Start with the given measurement: [katex]\text{5 km}[/katex]

  2. Use the conversion factors for kilometers to meters and meters to millimeters: [katex]\text{5 km} \times \frac{10^3 \, \text{m}}{1 \, \text{km}} \times \frac{10^3 \, \text{mm}}{1 \, \text{m}}[/katex]

  3. Perform the multiplication: [katex]\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}[/katex]

  4. Simplify to get the final answer: [katex]\boxed{5 \times 10^6 \, \text{mm}}[/katex]

Prefix

Symbol

Power of Ten

Equivalent

Pico-

p

[katex]10^{-12}[/katex]

Nano-

n

[katex]10^{-9}[/katex]

Micro-

µ

[katex]10^{-6}[/katex]

Milli-

m

[katex]10^{-3}[/katex]

Centi-

c

[katex]10^{-2}[/katex]

Deci-

d

[katex]10^{-1}[/katex]

(Base unit)

[katex]10^{0}[/katex]

Deca- or Deka-

da

[katex]10^{1}[/katex]

Hecto-

h

[katex]10^{2}[/katex]

Kilo-

k

[katex]10^{3}[/katex]

Mega-

M

[katex]10^{6}[/katex]

Giga-

G

[katex]10^{9}[/katex]

Tera-

T

[katex]10^{12}[/katex]

  1. 1. Some answers may vary by 1% due to rounding.
  2. Gravity values may differ: \(9.81 \, \text{m/s}^2\) or \(10 \, \text{m/s}^2\).
  3. Variables can be written differently. For example, initial velocity (\(v_i\)) may be \(u\), and displacement (\(\Delta x\)) may be \(s\).
  4. Bookmark questions you can’t solve to revisit them later
  5. 5. Seek help if you’re stuck. The sooner you understand, the better your chances on tests.

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