Kinematic Equations: Algebraic Method

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Core Concepts

$LaTeX: \definecolor{MyGreen}{RGB}{1,68,33} \textcolor{black}{v_{av}=\frac{\Delta x}{\Delta t}}\textcolor{MyGreen}{\Longrightarrow}$ $\definecolor{MyGreen}{RGB}{1,68,33} \textcolor{black}{v_{av}=\frac{\Delta x}{\Delta t}}\textcolor{MyGreen}{\Longrightarrow}$ Because velocity will always change uniformly with time (IF acceleration is a constant value, then $LaTeX: \definecolor{MyGreen}{RGB}{1,68,33} \textcolor{MyGreen}v_{av}$ $\definecolor{MyGreen}{RGB}{1,68,33} \textcolor{MyGreen}v_{av}$ will be the mean of both the initial and final velocities. $LaTeX: \definecolor{MyGreen}{RGB}{1,68,33} \textcolor{MyGreen}\Longrightarrow\textcolor{red}{ v_{av}=\frac{(v_{i}+v_{f})}{2}}$ $\definecolor{MyGreen}{RGB}{1,68,33} \textcolor{MyGreen}\Longrightarrow\textcolor{red}{ v_{av}=\frac{(v_{i}+v_{f})}{2}}$

$LaTeX: \definecolor{MyGreen}{RGB}{1,68,33} \textcolor{black}{a=\frac{\Delta v}{\Delta t}}\textcolor{MyGreen}{\Longrightarrow a}$ $\definecolor{MyGreen}{RGB}{1,68,33} \textcolor{black}{a=\frac{\Delta v}{\Delta t}}\textcolor{MyGreen}{\Longrightarrow a}$ is a constant value

Equation #1: $LaTeX: \textcolor{red}{\Delta x = {1 \over 2 } \left( v_i + v_f \right) t}$ $\textcolor{red}{\Delta x = {1 \over 2 } \left( v_i + v_f \right) t}$

Note: Depending on what textbook you are using, different subscripts may be used. A subscript "zero" (v₀) is the same as a subscript "i" (v_i).

$LaTeX: v_{av} = { \Delta x \over \Delta t }$ $v_{av} = { \Delta x \over \Delta t }$

$LaTeX: { \left( v_i + v_f \right) \over 2 } = { \left( x_f - x_i \right) \over \left( t_f - t_i \right) }$ ${ \left( v_i + v_f \right) \over 2 } = { \left( x_f - x_i \right) \over \left( t_f - t_i \right) }$ Cross multiply

$LaTeX: \left( v_i + v_f \right) \left( t_f - t_i \right) = 2 \left( x_f - x_i \right)$ $\left( v_i + v_f \right) \left( t_f - t_i \right) = 2 \left( x_f - x_i \right)$ Let t_i = 0 and t_f = t for simplicity

$LaTeX: \left( v_i + v_f \right) \left( t_f - 0 \right) = 2 \left( x_f - x_i \right)$ $\left( v_i + v_f \right) \left( t_f - 0 \right) = 2 \left( x_f - x_i \right)$ Substitute $LaTeX: \left( x_f - x_i \right) = \Delta x$ $\left( x_f - x_i \right) = \Delta x$

$LaTeX: \left( v_i + v_f \right) t = 2 \left( \Delta x \right)$ $\left( v_i + v_f \right) t = 2 \left( \Delta x \right)$ Divide both sides by 2 and rearrange.

$LaTeX: \textcolor{red}{\boxed{\Delta x = { 1 \over 2 } \left( v_i + v_f \right) t}}$ $\textcolor{red}{\boxed{\Delta x = { 1 \over 2 } \left( v_i + v_f \right) t}}$

Equation #2: $LaTeX: \textcolor{red}{v_f = v_i + at}$ $\textcolor{red}{v_f = v_i + at}$

$LaTeX: a = { \Delta v \over \Delta t }$ $a = { \Delta v \over \Delta t }$ Remember, "a" is a constant.

$LaTeX: a = { \left( v_f - v_i \right) \over \left( t_f - t_i \right) }$ $a = { \left( v_f - v_i \right) \over \left( t_f - t_i \right) }$ Left t_i = 0 and t_f = t.

$LaTeX: a = { v_f - v_i \over t - 0 }$ $a = { v_f - v_i \over t - 0 }$ Multiply both sides by t

LaTeX: a t = v_f - v_i $a t = v_f - v_i$ Solve for v_f

$LaTeX: \textcolor{red}{\boxed{v_f = v_i + a t}}$ $\textcolor{red}{\boxed{v_f = v_i + a t}}$

Equation #3: $LaTeX: \textcolor{red}{\Delta x = v_i t + {1 \over 2 } a t^2}$ $\textcolor{red}{\Delta x = v_i t + {1 \over 2 } a t^2}$

Note: This equation does not have a v_f term!

Start with the two equations we just derived:

1. $LaTeX: \Delta x = { 1 \over 2 } \left( v_i + v_f \right) t$ $\Delta x = { 1 \over 2 } \left( v_i + v_f \right) t$

2. LaTeX: v_f = v_i + a t $v_f = v_i + a t$

Substitute equation #2 into equation #1

$LaTeX: \Delta x = { 1 \over 2 } \left( v_i + v_i + a t \right) t$ $\Delta x = { 1 \over 2 } \left( v_i + v_i + a t \right) t$

$LaTeX: \Delta x = { 1 \over 2 } t \space \left( 2 v_i + a t \right)$ $\Delta x = { 1 \over 2 } t \space \left( 2 v_i + a t \right)$ Distribute the $LaTeX: { 1 \over 2 } t$ ${ 1 \over 2 } t$

$LaTeX: \Delta x = {1 \over 2 } t \left( 2 v_i \right)+ {1 \over 2 } t \left( a t \right)$ $\Delta x = {1 \over 2 } t \left( 2 v_i \right)+ {1 \over 2 } t \left( a t \right)$

$LaTeX: \textcolor{red}{\boxed{\Delta x=v_it+\frac{1}{2}at^2}}$ $\textcolor{red}{\boxed{\Delta x=v_it+\frac{1}{2}at^2}}$

Equation #4: $LaTeX: \textcolor{red}{v_f^2 = v_i ^2 +2a\Delta x}$ $\textcolor{red}{v_f^2 = v_i ^2 +2a\Delta x}$

Note: This equation does not have a "t" term!

Start with the two equations:

1. $LaTeX: \Delta x=\frac{1}{2}(v_i+v_f)t$ $\Delta x=\frac{1}{2}(v_i+v_f)t$

2. LaTeX: v_f=v_i +at $v_f=v_i +at$

Start by solving equation 2 for t:

LaTeX: v_f=v_i+at $v_f=v_i+at$

LaTeX: v_f-v_i=at $v_f-v_i=at$

$LaTeX: \frac{v_f-v_i}{a}=t$ $\frac{v_f-v_i}{a}=t$

$LaTeX: \textcolor{black}{t=}\textcolor{blue}{\frac{v_f-v_i}{a}}$ $\textcolor{black}{t=}\textcolor{blue}{\frac{v_f-v_i}{a}}$

Substitute the result into equation 1:

$LaTeX: \Delta x=\frac{1}{2}(v_i+v_f)$ $\Delta x=\frac{1}{2}(v_i+v_f)$

$LaTeX: \textcolor{black}{\Delta x=\frac{1}{2}(v_i+v_f)}\textcolor{blue}{\frac{v_f+v_i}{a}}$ $\textcolor{black}{\Delta x=\frac{1}{2}(v_i+v_f)}\textcolor{blue}{\frac{v_f+v_i}{a}}$

$LaTeX: \textcolor{black}{\Delta x=\frac{1}{2}}{\textcolor{blue}{\underbrace{\textcolor{black}{(v_i+v_f)}}}}\textcolor{black}{\frac{v_f+v_i}{a}}$ $\textcolor{black}{\Delta x=\frac{1}{2}}{\textcolor{blue}{\underbrace{\textcolor{black}{(v_i+v_f)}}}}\textcolor{black}{\frac{v_f+v_i}{a}}$ Switch the order of LaTeX: v_i $v_i$ and LaTeX: v_f $v_f$

$LaTeX: \Delta x=\frac{1}{2a}(v_f+v_i)(v_f-v_i)$ $\Delta x=\frac{1}{2a}(v_f+v_i)(v_f-v_i)$ Multiply both sides by 2a

$LaTeX: (2a)\Delta x=\frac{1}{2a}(v_f+v_i)(v_f-v_i)(2a)$ $(2a)\Delta x=\frac{1}{2a}(v_f+v_i)(v_f-v_i)(2a)$

$LaTeX: 2a\Delta x=\underbrace{(v_f+v_i)(v_f-v_i)}$ $2a\Delta x=\underbrace{(v_f+v_i)(v_f-v_i)}$ This is a difference of squares

$LaTeX: 2a\Delta x=v_f^2-v_i^2$ $2a\Delta x=v_f^2-v_i^2$

$LaTeX: \textcolor{red}{\boxed{v_f^2=v_i^2+2a\Delta x}}$ $\textcolor{red}{\boxed{v_f^2=v_i^2+2a\Delta x}}$