Talk:Physics with Calculus/Mechanics/Motion in Two Dimensions

$$ v^2 \equiv \mathbf{v} \cdot \mathbf{v} $$

$$ v^2 \hat{\mathbf{a}} = v_0^2 (2 \hat{\mathbf{v_0}} \cdot \hat{\mathbf{a}}  \hat{\mathbf{v_0}} - \hat{\mathbf{a}}) + 2 \| \mathbf{a} \| (\mathbf{r} - \mathbf{r_0} ) + 2 \| \mathbf{v} \| \| \mathbf{v_0} \| (\hat{\mathbf{v}} \cdot \hat{\mathbf{v_0}} \hat{\mathbf{a}} - \hat{\mathbf{v}} \cdot \hat{\mathbf{a}} \hat{\mathbf{v_0}}).$$

$$ v^2 \hat{\mathbf{a}} = v_0^2 (2 \hat{\mathbf{v_0}} \cdot \hat{\mathbf{a}}  \hat{\mathbf{v_0}} - \hat{\mathbf{a}}) + 2 \| \mathbf{a} \| (\mathbf{r} - \mathbf{r_0} ) $$

Untitled
Because (at least from what i think is going on) $$ (\hat{\mathbf{v}} \cdot \hat{\mathbf{v_0}} \hat{\mathbf{a}} - \hat{\mathbf{v}} \cdot \hat{\mathbf{a}} \hat{\mathbf{v_0}}) $$ is 0, because of the commutativity of the dot product.

$$ v^2 \hat{\mathbf{a}} = v_0^2 \hat{\mathbf{a}} (2 \hat{\mathbf{v_0}} \cdot  \hat{\mathbf{v_0}} - 1) + 2 \| \mathbf{a} \| (\mathbf{r} - \mathbf{r_0} ) $$ By the distributive property of the dot product.

But $$ \hat{\mathbf{v_0}} \cdot \hat{\mathbf{v_0}} $$ is 1, since $$ \hat{\mathbf{v_0}} $$ is a unit vector.

$$ v^2 \hat{\mathbf{a}} = v_0^2 \hat{\mathbf{a}} (2 - 1) + 2 \| \mathbf{a} \| (\mathbf{r} - \mathbf{r_0} ) $$

$$ v^2 \hat{\mathbf{a}} = v_0^2 \hat{\mathbf{a}} + 2 \| \mathbf{a} \| (\mathbf{r} - \mathbf{r_0} ) $$

By definition, $$ \hat{\mathbf{a}} = \frac{\mathbf{a}}{\| \mathbf{a} \|} $$

so

$$ v^2 \frac{\mathbf{a}}{\| \mathbf{a} \|} = v_0^2 \frac{\mathbf{a}}{\| \mathbf{a} \|} + 2 \| \mathbf{a} \| (\mathbf{r} - \mathbf{r_0} ) $$

Multiplying through by $$\|\mathbf{a}\|$$:

$$ v^2 \mathbf{a} = v_0^2 \mathbf{a} + 2 \| \mathbf{a} \| \| \mathbf{a} \| (\mathbf{r} - \mathbf{r_0} ) $$

Assuming we have defined $$ \| \mathbf{x} \| $$ as $$\sqrt{\mathbf{x} \cdot \mathbf{x}} $$:

$$ v^2 \mathbf{a} = v_0^2 \mathbf{a} + 2  \mathbf{a} \cdot \mathbf{a} (\mathbf{r} - \mathbf{r_0} ) $$

which is:

$$  \mathbf{a} \cdot (v^2  - v_0^2 - 2  \mathbf{a} \cdot (\mathbf{r} - \mathbf{r_0} ))  = 0$$

Since the stuff in the parens is a scalar, either the acceleration is 0 or the stuff in the parens is. Assuming a nonzero acceleration:

$$ \mathbf{a} \neq \mathbf{0} $$

than

$$ (v^2 - v_0^2 - 2  \mathbf{a}  \cdot (\mathbf{r} - \mathbf{r_0} ))  = 0 $$

which implies:

$$ v^2 = v_0^2 + 2  \mathbf{a} \cdot  (\mathbf{r} - \mathbf{r_0} )) $$

Which is _not_ useless, since this equation is not dependant on time. Thegeneralguy (talk) 15:35, 1 September 2008 (UTC)

"$$\Delta x = v_{xf}\Delta t + \begin{matrix} \frac{1}{2} \end{matrix} (v_{xf}-v_{xi}) \Delta t$$

This simplifies to:

$$\Delta x = (v_{xi} + \begin{matrix} \frac{1}{2} \end{matrix} v_{xf} - \begin{matrix} \frac{1}{2} \end{matrix} v_{xi})\Delta t$$"

Is this right? Does $$v_{xf}$$ simplify to $$v_{xi}$$?

I changed it to $$v_{xi}$$