Constant Acceleration Model | Primer on Geometric Algebra | David Hestenes

## Constant Acceleration Model

Free Particle Model.

Next sectionCircular Motion Model.

Constant acceleration model (without coordinates!!):

$\Large{\frac{{d{\bf{v}}}}{{dt}}\,\,\, = \,\,\,{\bf{g}}\,\,\,\,\,\, \Rightarrow \,\,\,\,\,\,{\bf{v}}\,\, = \,\,\,{{\bf{v}}_0}\,\, + \,\,{\bf{g}}t\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\frac{{d{\bf{x}}}}{{dt}}\,\,\, = \,\,\,{\bf{v}}\,\,\,\,\,\, \Rightarrow \,\,\,\,\,\,{\bf{x}}\,\,\, = \,\,\,{{\bf{x}}_0}\,\, + \,\,{{\bf{v}}_0}t\,\, + \,\,\frac{1}{2}{\bf{g}}{t^2}}$

(hodograph)                                      (trajectory)

Vector algebraic model
(where
r = range = x - x0):

$\Large{\bbox[8px,border:2px solid red]{\frac{1}{2}\left( {{\bf{v}} + {{\bf{v}}_0}} \right)\,\,\, = \,\,\,\frac{{\bf{r}}}{t}\,\,\, \equiv \,\,\,{\bf{\bar v}}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,{\bf{v}} - {{\bf{v}}_0}\,\,\, = \,\,\,{\bf{g}}t}}$

Reduces all projectile problems to "
solving a parallelogram!!"

Problem: Determine

(a)  the range r of a target sighted in a direction  r/r  that has been hit by a projectile,
launched with velocity
v0,

(b)  the launching angle for maximum range,

(c)  the time of flight.

General case: Elevated target.

Complicated solution with rectangular coordinates published in AJP.

Much simpler GA solution in NFCM (see references in Prolog).

(Click here for solutions for this problem.)

Simplest case: r is horizontal.

Hint: Consider (v0)(v0)

Solving a parallogram with GA:          − v0 gt                    v0 2r/t

We eliminate t by multiplication:    (v + v0)(v - v0= 2rg

Expanding:    v2 − v0v0v − vv0 v2 − v0+ 2v0v = 2(rrg

Separately equating scalar and bivector parts, we get:

v2 − v02 2rg                    v0v rg

The first equation may be recognized as conservation of energy (loss in KE = gain in PE).

The second equation represents a geometry theorem:

If a parallelogram has area A,
a new parallelogram, made with the diagonals of the first, has area 2
A.

For r horizontal, we have

$\Large{\begin{array}{l}{\bf{r}} \cdot {\bf{g}}\,\,\, = \,\,\,0\,\,\,\,\,\,\, \Rightarrow \,\,\,\,\,\,\,{v^2}\,\,\, = \,\,\,{v_0}^2\,\,\,\,\,\, \Rightarrow \\\left| {{{\bf{v}}_0} \wedge {\bf{v}}} \right|\,\,\, = \,\,\,{v_0}^2\,\,\sin \left( {2\theta } \right)\,\,\, = \,\,\,\left| {{\bf{r}} \wedge {\bf{g}}} \right|\,\,\, = \,\,\,rg\end{array}}$

Hence, the
horizontal range formula:

$\Large{\bbox[8px,border:2px solid red]{\begin{array}{l}r\,\,\, = \,\,\,\frac{{{v_0}^2}}{g}\sin \left( {2\theta } \right)\end{array}}}$

The general GA solution for an elevated target and time of flight is derived in
NFCM.

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Next section: Circular Motion Model.