PX153 - A3 - position, velocity and acceleration vectors

position vector, usually $\vec{r}$ , describes the position of point P relative to the origin
eg: for trajectory of a particle, $\underset{―}{r} (t) = x (t) \underset{―}{\hat{i}} + y (t) \underset{―}{\hat{i} j} + z (t) \underset{―}{\hat{k}}$

\underset{―}{v} (t) = \frac{d \underset{―}{r}}{d t} = \frac{d x (t)}{d t} \underset{―}{\hat{i}} + \frac{d y (t)}{d t} \underset{―}{\hat{j}} + \frac{d z (t)}{d t} \underset{―}{\hat{k}}

\underset{―}{a} (t) = \frac{d^{2} \underset{―}{r}}{d t^{2}} = \frac{d^{2} x (t)}{d t} \underset{―}{\hat{i}} + \frac{d^{2} y (t)}{d t^{2}} \underset{―}{\hat{j}} + \frac{d^{2} z (t)}{d t^{2}} \underset{―}{\hat{k}}

- cartesian coordinates, initial position ${\underset{―}{r}}_{o} = h_{o} \underset{―}{\hat{j}}$
- gravity is in the negative y direction $\underset{―}{g} = - g \underset{―}{\hat{j}}$
- initial velocity ( $v_{0}) = (v_{0 x}, v_{0 y}) = v_{0 x} \underset{―}{\hat{i}} + v_{0 y} \underset{―}{\hat{j}}$

equations of motion in vector notation

\vec{F} = m \vec{a} = m \vec{g}

\vec{a} = \frac{d^{2} r}{d t^{2}} = - g \underset{―}{\hat{j}} = \frac{d \vec{v}}{d t}

integrate to find velocity:

\vec{v} (t) = {\vec{v}}_{0} + \int_{0}^{t} \frac{d \vec{v}}{d t} d t = {\vec{v}}_{0} + \int_{0}^{t} - g \underset{―}{\hat{j}} . d t

\vec{v} = {\vec{v}}_{0} - g t \underset{―}{\hat{j}}

integrating again:

\vec{r} (t) = {\vec{r}}_{0} + \int_{0}^{t} \vec{v} (t) . d t = {\vec{r}}_{0} + {\vec{v}}_{0} t - \frac{1}{2} g t^{2} \underset{―}{\hat{j}}

the components:
- $x (t) = v_{0 x} t$
- $y (t) = h_{0} + v_{0 y} t - \frac{1}{2} g t^{2}$
  (check that $\vec{r} (0)$ and $\vec{v} (0)$ are correct)
  we can solve for $y (x)$ using $t = \frac{x}{v_{0 x}}$

y (x) = h_{0} + (\frac{v_{0 y}}{v_{0 x}}) x - \frac{1}{2} g \frac{x^{2}}{v_{0 x}^{2}}

a parabola as expected