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I'm trying to solve the following equation for a, my work is as follows.

$d_1=\frac{d_2-\frac{1}{2}at_2^2}{t_2}t_1+\frac{1}{2}at_1^2$

$d_1=\frac{d_2t_1-\frac{1}{2}at_2^2t_1}{t_2}+\frac{1}{2}at_1^2$

$d_1-\frac{d_2t_1}{t_2}=\frac{-\frac{1}{2}at_2^2t_1}{t_2}+\frac{1}{2}at_1^2$

$d_1-\frac{d_2t_1}{t_2}=-\frac{1}{2}at_2t_1+\frac{1}{2}at_1^2$

$d_1=-\frac{1}{2}at_2t_1+\frac{1}{2}at_1^2+\frac{d_2t_1}{t_2}$

$d_1=t_1(-\frac{1}{2}at_2+\frac{1}{2}at_1+\frac{d_2}{t_2})$

$\frac{d_1}{t_1}=-\frac{1}{2}at_2+\frac{1}{2}at_1+\frac{d_2}{t_2}$

$\frac{d_1}{t_1}-\frac{d_2}{t_2}=-\frac{1}{2}at_2+\frac{1}{2}at_1$

$\frac{d_1}{t_1}-\frac{d_2}{t_2}=a(-\frac{1}{2}t_2+\frac{1}{2}t_1)$

$a=\frac{\frac{d_1}{t_1}-\frac{d_2}{t_2}}{-\frac{1}{2}t_2+\frac{1}{2}t_1}$

As you can see I end up with a really messy fraction at the end. Can I rewrite the above as the following?

$a=\frac{d_1-d_2}{t_1t_2(-\frac{1}{2}t_2+\frac{1}{2}t_1)}$

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    $\begingroup$ You forgot $t$s in the numerator. The best expression is $a={2 \over t_1 t_2}{d_1t_2-d_2t_1 \over t_1-t_2}$ $\endgroup$ – N74 Sep 20 '18 at 8:54
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No. You have proceeded correctly thus far. We have $$a = \frac{\frac{d_1}{t_1} - \frac{d_2}{t_2}}{-\frac{t_2}{2}+\frac{t_1}{2}}$$

Consider the numerator: $$\frac{d_1}{t_1} - \frac{d_2}{t_2} = \frac{d_1t_2-d_2t_1}{t_1t_2}$$

The denominator is: $$\frac{t_1-t_2}{2}$$

Now simplify the expression to get: $$a = \frac{2[d_1t_2-d_2t_1]}{t_1t_2[t_1-t_2]}$$

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  • $\begingroup$ Hm, the above simplification actually breaks the relationship of my problem. I know this isn't relevant to my question but this is a physics problem and a is an acceleration. Thus I'm expecting an answer to be in the form of d/t^2. $\endgroup$ – Renari Sep 20 '18 at 8:34

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