AP Physics

Unit 1 - Vectors and Kinematics

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Step Derivation/Formula Reasoning
1 Calculate the car’s speed after acceleration:

Initial speed: \( u_0 = 0 \, \text{m/s} \)
Acceleration: \( a_1 = 3.5 \, \text{m/s}^2 \)
Time: \( t_1 = 10 \, \text{s} \)

\( v_1 = u_0 + a_1 t_1 \)
\( v_1 = 0 + (3.5)(10) = 35 \, \text{m/s} \)

Determined the speed after accelerating for 10 seconds.
2 Calculate distance during acceleration:

\( s_1 = u_0 t_1 + \tfrac{1}{2} a_1 t_1^2 \)
\( s_1 = 0 + \tfrac{1}{2} (3.5)(10)^2 = 175 \, \text{m} \)

Found distance covered during acceleration phase.
3 Determine distance during reaction time:

Reaction time: \( t_r = 0.6 \, \text{s} \)
\( s_{\text{react}} = v_1 t_r \)
\( s_{\text{react}} = (35)(0.6) = 21 \, \text{m} \)

Calculated distance traveled during driver’s reaction time.
4 Calculate remaining distance to the ramp:

Distance ahead when noticing ramp: \( 50 \, \text{m} \)
Remaining distance after reaction: \( d_{\text{remain}} = 50 – 21 = 29 \, \text{m} \)

Determined how much distance is left to brake.
5 Compute stopping distance required:

Braking acceleration: \( a_2 = -7.2 \, \text{m/s}^2 \)
\( v^2 = u^2 + 2 a s \)
\( 0 = (35)^2 + 2(-7.2) s_{\text{brake}} \)
\( s_{\text{brake}} = \dfrac{(35)^2}{2 \times 7.2} \approx 85.07 \, \text{m} \)

Calculated distance needed to stop completely.
6 Find speed at the ramp:

\( v^2 = u^2 + 2 a s \)
\( v^2 = (35)^2 + 2(-7.2)(29) \)
\( v^2 = 1225 – 417.6 = 807.4 \)
\( v = \sqrt{807.4} \approx 28.43 \, \text{m/s} \)

Determined the car’s speed upon reaching the ramp.
7 Resolve velocity into components at the ramp:

Ramp angle: \( \theta = 27^\circ \)
\( v_x = v \cos \theta = (28.43) \cos 27^\circ \approx 25.34 \, \text{m/s} \)
\( v_y = v \sin \theta = (28.43) \sin 27^\circ \approx 12.91 \, \text{m/s} \)

Found horizontal and vertical components of velocity.
8 (a) Calculate time of flight after ramp:

Vertical motion equation:
\( y = v_y t – \tfrac{1}{2} g t^2 \)
Height difference: \( y = -3 \, \text{m} \)
\( -3 = (12.91) t – 4.9 t^2 \)
Rearranged to quadratic: \( 4.9 t^2 – 12.91 t – 3 = 0 \)
Solved for \( t \):
\( t = \dfrac{12.91 \pm \sqrt{12.91^2 – 4 \times 4.9 \times (-3)}}{2 \times 4.9} \)
\( t \approx 2.85 \, \text{s} \)

Calculated time the car is airborne after the ramp.
9 Compute horizontal distance traveled:

\( \Delta x = v_x t \)
\( \Delta x = (25.34)(2.85) \approx 72.22 \, \text{m} \)

Found horizontal distance after going off the ramp.
10 (b) Determine final vertical velocity:

\( v_{y_{\text{final}}} = v_y – g t \)
\( v_{y_{\text{final}}} = 12.91 – (9.8)(2.85) \approx -15.02 \, \text{m/s} \)

Calculated vertical component of velocity upon landing.
11 Calculate final speed and direction:

\( v_{\text{final}} = \sqrt{v_x^2 + v_{y_{\text{final}}}^2} \)
\( v_{\text{final}} = \sqrt{(25.34)^2 + (-15.02)^2} \approx 29.46 \, \text{m/s} \)

Direction angle:
\( \theta_{\text{final}} = \arctan\left( \dfrac{|v_{y_{\text{final}}}|}{v_x} \right) \)
\( \theta_{\text{final}} = \arctan\left( \dfrac{15.02}{25.34} \right) \approx 30.7^\circ \) below horizontal

Found the magnitude and angle of the car’s velocity upon impact.

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(a) The horizontal distance the car travels after going off the ramp is approximately \( 72.2 \, \text{m} \).

(b) The car’s final velocity upon reaching the ground is approximately \( 29.5 \, \text{m/s} \) at an angle of \( 30.7^\circ \) below the horizontal.

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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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