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

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

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

$v^2 \hat{\mathbf{a}} = v_0^2 (2 \hat{\mathbf{v_0}} \cdot \hat{\mathbf{a}} \hat{\mathbf{v_0}} - \hat{\mathbf{a}}) + 2 \| \mathbf{a} \| (\mathbf{r} - \mathbf{r_0} )$ 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 $v x f$ simplify to $v x i$ ?

I changed it to $v x i$

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

From Wikibooks, the open-content textbooks collection

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