AP Physics

Unit 1 - Vectors and Kinematics

Intermediate

Mathematical

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Step Derivation/Formula Reasoning
(a) How fast is the rocket traveling when the engine cuts off?
1 Use the kinematic equation:

\( v^2 = u^2 + 2 a s \)
Initial speed \( u = 0 \, \text{m/s} \), acceleration \( a = 12.0 \, \text{m/s}^2 \), distance \( s = 1,000 \, \text{m} \).

Relates final velocity, initial velocity, acceleration, and distance.
2 Substitute values:

\( v^2 = 0 + 2 \times 12.0 \times 1,000 \)
\( v^2 = 24,000 \, \text{m}^2/\text{s}^2 \)

Calculated \( v^2 \) using given values.
3 Solve for \( v \):

\( v = \sqrt{24,000} \approx 154.92 \, \text{m/s} \)

Found the rocket’s speed at engine cutoff.
(b) What maximum height relative to the ground does the rocket reach?
4 After engine cutoff, use \( v^2 = u^2 + 2 a s \) with \( v = 0 \, \text{m/s} \), \( u = 154.92 \, \text{m/s} \), \( a = -9.8 \, \text{m/s}^2 \):

\( 0 = (154.92)^2 + 2 (-9.8) s \)

Used kinematic equation for upward motion until velocity is zero.
5 Solve for \( s \):

\( s = \dfrac{(154.92)^2}{2 \times 9.8} \)
\( s = \dfrac{24,000}{19.6} \approx 1,224.49 \, \text{m} \)

Calculated additional height after engine cutoff.
6 Total maximum height:

\( h_{\text{total}} = 1,000 + 1,224.49 = 2,224.49 \, \text{m} \)

Added initial altitude to additional height for total height.
(c) Velocity just before the rocket hits the earth.
7 Use \( v^2 = u^2 + 2 a s \) for free fall from maximum height with \( u = 0 \, \text{m/s} \), \( a = 9.8 \, \text{m/s}^2 \), \( s = 2,224.49 \, \text{m} \):

\( v^2 = 0 + 2 \times 9.8 \times 2,224.49 \)

Calculated final velocity during descent.
8 Compute \( v^2 \):

\( v^2 = 43,598.00 \, \text{m}^2/\text{s}^2 \)
\( v = \sqrt{43,598.00} \approx 208.71 \, \text{m/s} \)

Found velocity just before impact.
(d) Total amount of time the rocket was in the air.
9 **Time during powered ascent:**

Use \( s = u t + \dfrac{1}{2} a t^2 \) with \( s = 1,000 \, \text{m} \), \( u = 0 \, \text{m/s} \), \( a = 12.0 \, \text{m/s}^2 \):
\( 1,000 = 0 + 6.0 t^2 \)

Calculated time for powered ascent.
10 Solve for \( t \):

\( t^2 = \dfrac{1,000}{6.0} \approx 166.67 \)
\( t_{\text{ascent}} = \sqrt{166.67} \approx 12.91 \, \text{s} \)

Found time during powered ascent.
11 **Time during coasting ascent:**

Use \( v = u + a t \) with \( v = 0 \, \text{m/s} \), \( u = 154.92 \, \text{m/s} \), \( a = -9.8 \, \text{m/s}^2 \):
\( 0 = 154.92 – 9.8 t \)

Calculated time from engine cutoff to maximum height.
12 Solve for \( t \):

\( t = \dfrac{154.92}{9.8} \approx 15.81 \, \text{s} \)

Found time during coasting ascent.
13 **Time during free fall descent:**

Use \( s = \dfrac{1}{2} a t^2 \) with \( s = 2,224.49 \, \text{m} \), \( a = 9.8 \, \text{m/s}^2 \):
\( 2,224.49 = 4.9 t^2 \)

Calculated time for descent back to earth.
14 Solve for \( t \):

\( t^2 = \dfrac{2,224.49}{4.9} \approx 454.9996 \)
\( t_{\text{descent}} = \sqrt{454.9996} \approx 21.21 \, \text{s} \)

Found time during free fall descent.
15 **Total time in the air:**

\( t_{\text{total}} = t_{\text{ascent}} + t_{\text{coasting}} + t_{\text{descent}} \)
\( t_{\text{total}} = 12.91 + 15.81 + 21.21 \approx 49.93 \, \text{s} \)

Summed all time intervals for total flight time.

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  1. \( 154.92 \, \text{m/s} \)
  2. \( 2,224.49 \, \text{m} \)
  3. \( 208.71 \, \text{m/s} \) downwards
  4. \( 49.93 \, \text{s} \)

Note: Answers may be off by \( \pm 0.2 \).

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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]

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