This page is designed to help you learn to solve problems more easily by getting you into the mindset of mixing and matching formulas. Note: This is a draft in progress at the moment, but I decided to hit “publish” anyway. If you spot any errors or terrible typos, please let me know. I still have a lot to add.

Want to jump straight to a specific variable?

  1. Acceleration
  2. Force
  3. Energy and Work

Quick Review of Vector Components

(using Force as the example, and assuming that θ is measured from the positive x-axis)

Components \\ \; \\ \Large \mathrm{F_x = Fcos\theta} \\ \; \\ \: \mathrm{F_y = Fsin\theta}
Resultant   \\ \; \\ \Large \mathrm{F = \sqrt{F_x^2 + F_y^2}}  \\ \; \\ \mathrm{\theta = tan^{-1}(\frac{F_y}{F_x})}

Kinematics

(constant acceleration)

 \textbf{Horizontal Motion} \\ \: \\ \Large \mathrm{a_x = constant}  \\ \: \\ \: \mathrm{x = \frac{1}{2}a_xt^2 + v_{o_x}t + x_o} \\ \: \\ \: \mathrm{v_x = v_{o_x} + a_xt} \\ \: \\ \: \mathrm{{v_x}^2 = {v_{o_x}}^2 + 2a_x(x - x_o)} \\ \: \\ \: \\ \textbf{Vertical Motion (Free Fall)} \\ \: \\ \Large \mathrm{a_y = g = constant} \\ \: \\ \mathrm{y = \frac{1}{2}gt^2 + v_{o_y}t + y_o} \\ \: \\ \: \mathrm{v_y = v_{o_y} + gt} \\ \: \\ \mathrm{{v_y}^2 = {v_{o_y}}^2 + 2g(y - y_o)} \\ \: \\Note: If your positive axis points upward, remember to include a negative sign when plugging in the value of g in these equations!  \\ \: \\ \: \\ \textbf{For Projectile Motion} \\ \: \\ \Large \mathrm{a_x = 0 [\frac{m}{s^2}]} \\ \: \\ \: \mathrm{v_x = v_{o_x}}  \\ \: \\  \normalsize \text{and the Range Formula is} \\ \: \\ \: \Large \mathrm{R = \frac{{v_o}^2 \sin{2\theta_o}}{g}} \\ \:
 \textbf{Angular Motion} \\ \: \\ \Large \mathrm{\alpha = constant} \\ \: \\ \mathrm{\theta = \frac{1}{2}\alpha t^2 + \omega_ot + \theta_o} \\ \: \\ \: \mathrm{\omega = \omega_o + \alpha t} \\ \: \\ \mathrm{\omega^2 = {\omega_o}^2 + 2\alpha(\theta - \theta_o)} \\ \: \\

Distance & Displacement

\textbf{Linear Distance}  \\ \: \\\Large \mathrm{Δx =} Total Length Traveled 

Distance is a scalar – direction does NOT matter.

In other words, a round trip that retraces the original path on the trip home will have a total distance that is twice the length of the one-way trip.

\\ \: \\ \: \\ \textbf{Linear Displacement} \\ \: \\\Large \mathrm{\Delta \vec{x} = x - x_o} \\ \: \\ \mathrm{\Delta \vec{y} = y - y_o} \\ \: \\

Displacement is a vector – direction matters.
Displacement is the shortest distance between the starting and ending positions.

For example, for the same round trip in which your return journey retraces the original path so you wind up back where you started, your total displacement is zero.

\textbf{Arc Length} \\ \: \\ \Large l \mathrm{=r \theta} \\ where r is the radius of the circle in \textbf{radians} and l is the length of the arc traveled on the circumference of the circle. Note: This equation is only useful for angles less than 15°. \\ \: \\ \textbf{Angular Displacement}  \\ \: \\ \Large \mathrm{\Delta \theta = \theta - \theta_o = }  the distance in degrees traveled around the circle. \\ \; \\   

Velocity

Units: \: \text{Linear:} \, \Large \mathrm{[\frac{m}{s}],} \: \: \normalsize \text{Angular:} \, \Large \mathrm{[\frac{rad}{s}]}
\textbf{Average Linear Velocity} \\ \Large \mathrm{\bar{v} = \frac{\Delta \vec{r}}{\Delta t}} \\ \: \\ \mathrm{\bar{v}_x = \frac{x - x_0}{\Delta t}} \\ \: \\ \mathrm{\bar{v}_y= \frac{y - y_0}{\Delta t}} \\ \: \\ \textbf{Instantaneous Linear Velocity} \Large \\ \mathrm{\vec{v} = \frac{d \vec{r}}{dt} = \frac{dx}{dt}\hat{i} + \frac{dy}{dt}\hat{j} + \frac{dz}{dt}\hat{k}} (Calculus)
\textbf{Average Angular Velocity} \\ \Large \mathrm{\bar{\omega} = \frac{\Delta \theta}{\Delta t} = \frac{\theta-\theta_0}{{\Delta t}}} \\ \: \\ \textbf{Instantaneous Angular Velocity} \\  \Large \mathrm{\omega = \frac{d \theta}{dt}} (Calculus)

Acceleration

Units: \Large \mathrm{[\frac{m}{s^2}]}
\Large \mathrm{\vec{a}=\frac{\Delta \vec{v}}{\Delta t}} \\ (algebra & trig)
\Large \mathrm{\vec{a}=\frac{d \vec{v}}{dt} = \frac{d^2 \vec{r}}{dt^2}} \\ (Calculus)
\Large \mathrm {\vec{a}=\frac{\vec{F}}{m}}
\Large \mathrm{a_{rad} = a_\perp = \frac{v^2}{r}}
\Large \mathrm {a_{tan} = \frac{\Delta v_{tan}}{\Delta t}}
\Large \mathrm{a_{tan}= r \frac{ \Delta \omega}{\Delta t}}
\Large \mathrm{a_{tan} = r \alpha}

Force

Units: [newtons] or [N] or \Large \mathrm{[\frac{kg·m}{s^2}]}
\Large \mathrm {\vec{F}={{m}\vec{a}}}
\Large \mathrm{\vec{F} = \frac{GMm}{\vec{r^2}}}
\Large \mathrm{F_{radial} = F_\perp= ma_\perp = \frac{mv^2}{r}}
\Large \mathrm {\vec{F}={{q}\vec{E}}}
\Large \mathrm{\vec{F} = \frac{kq_1q_2}{\vec{r^2}}}
\Large \mathrm {\vec{F}={{qv}\vec{B}sin(\theta)}}
\Large \mathrm {\vec{F}_{max}={{qv}\vec{B}}}
\Large \mathrm {\vec{F}_{max}={I}} l \mathrm{\vec{B}}

where l is the length of straight wire carrying current I in a magnetic field B and the angle θ between the magnetic field and the direction of the current is 90º

\Large \mathrm {\vec{F}={I}} l \mathrm {\vec{B}sin(\theta)}

where l is the length of straight wire carrying current I and θ is the angle between the magnetic field and the direction of the current

\Large \mathrm{\vec{F} = } \frac{\mathrm{\mu_0I_1I_2}l}{\mathrm {2\pi d}}

for the force between 2 current-carrying wires of length l a distance d apart

Energy and Work

\large \mathrm{Units: [joules]} or \large \mathrm{[J]} or \large \mathrm{[Nm]} or \large \mathrm{[CV]} or \large \mathrm{[FV^2]} or \Large \mathrm{[\frac{C^2}{F}]} \\ \hspace{1.2cm} \mathrm{also [eV] where 1[eV] = 1.6022x10^{-19}[J]}

Work and Change in Energy are the SAME THING!!!

Note: Work is only done by parallel forces!
Perpendicular forces are called centripetal; they do NO work.
They do NOT change the energy, so they do NOT change the speed.
Perpendicular forces can only change the DIRECTION of the velocity, not the magnitude.

\Large \mathrm{W = \Delta KE = -\Delta PE}
\Large \mathrm{W = F_\parallel d}
\Large \mathrm{W = -qEd}
\Large \mathrm{W = -qV_{ba}}

Conservation of Energy: Total Energy is Constant:
ΔKE = -ΔPE if there are no non-conservative forces present

\Large \mathrm{\Delta KE = \frac{1}{2}m{v_f}^2 - \frac{1}{2}m{v_0}^2 = W_{ba}}
\Large \mathrm{\Delta PE = mgh_f - mgh_0 = -W_{ba}}
\Large \mathrm{\Delta PE = qV_{ba} = -W_{ba}}

Energy due to an EM wave

\Large \mathrm{E = \frac{hc}{\lambda} = hf} \\ \: \\\mathrm{E = h\nu}

Momentum

\large \mathrm{Units: [\frac{kg·m}{s}]}
\Large \mathrm{\Delta \vec{p} = m\vec{v}_f - m\vec{v}_0 }