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Derivations of Introductory Physics Formulary

Posted Mar24-11 at 10:18 PM by jhae2.718
Updated May19-11 at 08:53 AM by jhae2.718
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Introduction
For reference, I thought I'd post some derivations of introductory calculus-based physics formulas, starting with those from mechanics.

Nomenclature
Bold face indicates a vector quantity.
Italic variables are scalar quantities.
For first and second order time derivatives, Newtonian notation is used.
For all other derivatives, Leibniz notation is used.

I. Kinematics
The most basic equation in kinematics is
$$\mathbf{a} = \mathbf{\dot{v}} = \mathbf{\ddot{r}}$$.
The first order equation can be solved to give the velocity as a function of time, and the second order equation can be solved to give the displacement as a function of time. The second order equation can also be rewritten as a first order equation in terms of t:
$$\mathbf{v} = \mathbf{\dot{r}}$$

Kinematics with Constant Acceleration
Velocity as a function of time
The initial value problem for velocity is as follows: given an constant acceleration a, and v(0)=v0, find v(t).
The solution is as follows:
\begin{align*} \mathbf{a} &= \mathbf{\dot{v}} \\ d\mathbf{v} &= \mathbf{a}dt \\ \mathbf{v} &= \int\!\!\mathbf{a}dt \\ \mathbf{v}(t) &= \mathbf{a}t + C \\ \text{From the IVP:}\\ \mathbf{v}(0) &= C = \mathbf{v}_0\\ \mathbf{v}(t)&=\mathbf{a}t+\mathbf{v}_0 \end{align*}

Position as a function of time
The initial value problem for displacement is as follows: given an constant acceleration a and v(t), find r(t) where the boundary condition is r(0)=r0.
The solution follows:
\begin{align*} \mathbf{v} &= \mathbf{ \dot{r} } \\ d \mathbf{r} &= \mathbf{v} dt \\ \mathbf{r} &= \int \! \! ( \mathbf{a} t + \mathbf{v}_0 ) dt \\ \mathbf{r} (t) &= \frac{1}{2} \mathbf{a} t^2 + \mathbf{v}_0 t + C \\ \text{From the IVP:} \\ \mathbf{r} (0) &= C = \mathbf{r}_0 \\ \mathbf{r} (t)&= \frac{1}{2} \mathbf{a} t^2 + \mathbf{v}_0 t + \mathbf{r}_0 \end{align*}

Kinematics Independent of Time
Recall the equation for the velocity as a function of time, $\mathbf{v}(t)=\mathbf{a}t+\mathbf{v}_0$. We can use this equation and the displacement equation to remove the time parameter and describe kinematics in terms of acceleration, velocity, and displacement.

First, solve for t in terms of a and v, letting v(t) = vf:
\begin{align*} \mathbf{v}_f&=\mathbf{a}t+\mathbf{v}_0 \\ t & = \frac{\mathbf{v}_f-\mathbf{v}_0}{\mathbf{a}} \end{align*}
Substitute this expression for t in r(t), letting r(t)-r0 = Δr:
\begin{align*} \Delta \mathbf{r} &= \frac{1}{2} \mathbf{a} t^2 + \mathbf{v}_0t \\ \Delta \mathbf{r} &= \frac{1}{2} \mathbf{a} \left( \frac{ \mathbf{v}_f - \mathbf{v}_0}{ \mathbf{a}} \right) ^2+ \mathbf{v}_0 \left( \frac{ \mathbf{v}_f - \mathbf{v}_0}{ \mathbf{a} } \right) \\ \Delta \mathbf{r} &= \frac{1}{2}\frac{\left(\mathbf{v}_f^2-2\mathbf{v}_f\mathbf{v}_0+\mathbf{v}_0^2\right)}{ \mathbf{a} }+\mathbf{v}_0\left(\frac{\mathbf{v}_f-\mathbf{v}_0}{ \mathbf{a}} \right) \\ \Delta \mathbf{r} &= \frac{\mathbf{v}_f^2-2\mathbf{v}_f\mathbf{v}_0+\mathbf{v}_0^2}{2\mathbf{a}}+\frac{2\mathbf{v }_0\mathbf{v}_f-2\mathbf{v}_0^2}{2 \mathbf{a}} \\ \Delta \mathbf{r} &= \frac{\mathbf{v}_f^2-\mathbf{v}_0^2}{2 \mathbf{a}} \\ \mathbf{v}_f^2-\mathbf{v}_0^2 &= 2\mathbf{a}\Delta\mathbf{r} \\ \mathbf{v}_f^2 &=\mathbf{v}_0^2 + 2\mathbf{a}\Delta\mathbf{r} \end{align*}

The Range Equation
The range equation gives the horizontal distance traveled by a projectile under the following assumptions:
• The only acceleration on the object is gravitational acceleration g.
• The object is given initial velocity v0 at angle θ.
• The initial and final positions occur at the same height. (I.e. this is a special case of a more general equation with a nontrivial derivation.)
• The initial displacement in the x-projection, x0, is zero.
• A standard right hand coordinate system is used with positive x to the right and positive y up.
From these assumptions, our equations of motion are:

x(t) = v0cos(θ)t
y(t) = v0sin(θ)t-1/2gt2

To solve for the range, we want the final time tf. To solve for this time, we find t such that y=0:
\begin{align*} 0 &= v_0\sin(\theta)t_f-\frac{gt_f^2}{2} \\ 0 &= t_f\left(v_0\sin(\theta)-\frac{gt_f}{2}\right) \\ \end{align*}
Here two possible solutions exist. Since we want the final time, tf=0 is an extraneous solution, leaving us with:
\begin{align*} 0 &= v_0\sin(\theta)-\frac{gt_f}{2} \\ v_0\sin(\theta)&=\frac{gt_f}{2} \\ t_f &= \frac{2v_0\sin(\theta)}{g} \\ \end{align*}
This result is the final time of a projectile.

We can substitute this equation into x(t) to find the range:
\begin{align*} x(t_f) &= v_0 \cos(\theta) \left( \frac{2 v_0 \sin( \theta ) }{g} \right) \\ x(t_f) &= \frac{v_0^2 \cdot 2 \sin( \theta ) \cos( \theta) }{g} \\ \text{Recall that } 2 \sin( \theta ) \cos( \theta ) = \sin( 2 \theta ): \\ x(t_f) &= \frac{v_0^2 \sin( 2 \theta )}{g} \\ \end{align*}
This is the familiar equation for projectile range.

Next up I'll post derivations for dynamics equations.
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1.  Corrected $\LaTeX$ markup to display properly under MathJax. Math looks best under HTML-CSS rendering. Right click on an equation, go to Settings > Math Renderer and select HTML-CSS. Posted May19-11 at 08:53 AM by jhae2.718