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$\def\b#1{\mathbf#1}$ Here's the equation where $ \b{a}$ and $t$ are unknown (all boldface variables are $3$ element vectors and $t$ is a scalar): $$ \tfrac12 \b{a}t^2 + \b{v_0}t + \b{p_0} = \tfrac12 \b{a_T}t^2 + \b{v_T}t + \b{p_T} $$

I know the magnitude I want for $\b{a}$: $$ \| \b{a} \| = K $$

where $K$ is known. Is there a way to use these to solve for $\b{a}$ and $t$ analytically?

We can reorganize the original equation as: $$ \b{a}t^2 = \b{a_T}t^2 + 2(\b{v_T}-\b{v_0})t + 2(\b{p_T}-\b{p_0}) $$ but I'm not sure where to go from there.

mentics
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  • To clarify: you know $\mathbf{b}$ and $\mathbf{c}$ and the magnitude $| \mathbf{a} |$? – Sammy Black May 19 '15 at 02:35
  • @taotree: $Kt^2=||\mathbf{b}t+\mathbf{c}||$ is surely solvable for $t$, then $\mathbf{a}=-\mathbf{b}\frac{1}{t}-\mathbf{c}\frac{1}{t^2}$ unless $t=0$. – Alexey Burdin May 19 '15 at 02:53
  • @SammyBlack yes. those are known. – mentics May 19 '15 at 03:24
  • I edited the question after some of the above comments. The question contained an incorrect assumption so the above comments use different variable names. But the comment did lead me to the solution. Thank you! – mentics May 22 '15 at 17:39

2 Answers2

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Assume that $b$ and $c$ are linearly independent. Then necessarily $a=\lambda b+\mu c$ for certain real $\lambda$, $\mu$. From $at^2+bt+c=0$ we then deduce $$\lambda t^2+t=0,\qquad \mu t^2+1=0\ .$$ It follows that $t\ne0$, so that we can write $\lambda=-{1\over t}$, $\>\mu=-{1\over t^2}$. The condition $|a|=K$ then enforces $${1\over t^2}|b|^2+{2\over t^3}b\cdot c+{1\over t^4}|c|^2=K^2\ ,$$ a fourth degree equation for $t$.

Christian Blatter
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  • That's the same equation I put in the update in the question. How does one solve that fourth degree equation? Oh... I think this is the answer: http://math.stackexchange.com/questions/785/is-there-a-general-formula-for-solving-4th-degree-equations – mentics May 19 '15 at 13:19
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Thanks for the help with comments. I'll try to write up the whole answer for the original equation with steps from the original equation. It wasn't quite as simple because converting to the general form was messier than anticipated because the unknown was "inside" the squared term.

So, we start with the original equation: $$ \tfrac12 \b{a}t^2 + \b{v_0}t + \b{p_0} = \tfrac12 \b{a_T}t^2 + \b{v_T}t + \b{p_T} $$ with some simple algebra becomes: $$ \b{a}t^2 = \b{a_T}t^2 + 2(\b{v_T}-\b{v_0})t + 2(\b{p_T}-\b{p_0}) $$

Substituting in $\b{b} = 2(\b{v_T}-\b{v_0})$, and $\b{c} = 2(\b{p_T}-\b{p_0})$, we get: $$ \b{a}t^2 = \b{a_T}t^2 + \b{b}t + \b{c} $$

Taking the magnitude of both sides: $$ \|\b{a}\|t^2 = \|\b{a_T}t^2 + \b{b}t + \b{c}\| $$

which can be converted to: $$ (\|\b{a}\|t^2)^2 = ({\bf a_T}\cdot{\bf a_T})t^4 + 2({\bf a_T}\cdot{\bf b})t^3 + ({\bf b}\cdot{\bf b} + 2({\bf a_T}\cdot{\bf c}))t^2 + 2({\bf b}\cdot{\bf c})t + ({\bf c}\cdot{\bf c}) $$

We're given $\|\b{a}\| = K$, so substituting that in and getting a 0 on the left gives us:

$$ 0 = ({\bf a_T}\cdot{\bf a_T}-K^2)t^4 + 2({\bf a_T}\cdot{\bf b})t^3 + ({\bf b}\cdot{\bf b} + 2({\bf a_T}\cdot{\bf c}))t^2 + 2({\bf b}\cdot{\bf c})t + ({\bf c}\cdot{\bf c}) $$

Now, we have a 4th degree polynomial in general form and can use techniques described here to solve for t. Then once we have t, we use the original equation to produce $\b{a}$:

$$ \b{a} = \b{a_T} + 2(\b{v_T}-\b{v_0})\frac1t + 2(\b{p_T}-\b{p_0})\frac1{t^2} $$

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