AP Physics

Unit 1 - Vectors and Kinematics

Intermediate

Mathematical

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Table 1: Determine Required Ground Speed

Step Derivation/Formula Reasoning
1 \( \Delta x = 450 \, \text{km} \) The total distance to the airport is \(450 \, \text{km}\).
2 \( t = 3.0 \, \text{h} \) The total time to reach the airport is \(3.0 \, \text{hours}\).
3 \( v_x = \frac{\Delta x}{t} \) The ground speed \( v_x \) required to reach the airport in the given time is the total distance divided by the total time.
4 \( v_x = \frac{450 \, \text{km}}{3.0 \, \text{h}} = 150 \, \text{km/h} \) By calculation, the required ground speed is \(150 \, \text{km/h}\) due south.

Table 2: Determine Required Airspeed and Heading

Step Derivation/Formula Reasoning
1 \( v_{\text{wind}} = 50.0 \, \text{km/h} \) The wind speed blowing from the west is \(50.0 \, \text{km/h}\).
2 \( v_x = 150 \, \text{km/h} \) The ground speed required to travel due south is \(150 \, \text{km/h}\).
3 \( v_{\text{plane}} \cos(\theta) = v_x \) Horizontal component of the plane’s airspeed must counteract the wind to achieve the required southward ground speed.
4 \( v_{\text{plane}} \sin(\theta) = v_{\text{wind}} \) Vertical component of the plane’s airspeed must counteract the wind speed blowing from the west.
5 \( v_{\text{plane}} = \sqrt{v_x^2 + v_{\text{wind}}^2} \) Use Pythagorean theorem to find the magnitude of the plane’s airspeed vector.
6 \( v_{\text{plane}} = \sqrt{(150 \, \text{km/h})^2 + (50.0 \, \text{km/h})^2} \) Substitute the known values into the equation.
7 \( v_{\text{plane}} = \sqrt{22500 \, \text{km}^2/\text{h}^2 + 2500 \, \text{km}^2/\text{h}^2} \) Simplify the square terms inside the square root.
8 \( v_{\text{plane}} = \sqrt{25000 \, \text{km}^2/\text{h}^2} = 158.1 \, \text{km/h} \) Evaluate the square root to find the plane’s airspeed.
9 \( \theta = \arctan\left(\frac{v_{\text{wind}}}{v_x}\right) \) Use the inverse tangent function to find the heading angle \( \theta \) east of south.
10 \( \theta = \arctan\left(\frac{50.0}{150}\right) \) Substitute the values for wind speed and ground speed.
11 \( \theta = \arctan\left(\frac{1}{3}\right) = 18.4^\circ \) Evaluate the inverse tangent to find the heading angle.
12 \( v_{\text{plane}} = 158.1 \, \text{km/h}, \theta = 18.4^\circ \text{ east of south} \) The required airspeed is \(158.1 \, \text{km/h}\) and the heading is \(18.4^\circ \) east of south.

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\( v_{\text{plane}} = 158.1 \, \text{km/h}, \theta = 18.4^\circ \text{ east of south} \).

Note you could have also used the angle South of East, which would have been \(\theta = 18.4^\circ \text{ south of east} \). Either direction is acceptable.

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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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