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I want to solve $t'$ in:

$$ut + \dfrac{at^2}{2} = ut' + \dfrac{a't'^2}{2}$$

Where I will know the values for constants $u$, $t$, $a$ and $a'$.

I believe I can reduce the above to a quadratic equation:

$$2ut + at^2 = 2ut' + a't'^2$$

$$a't'^2 + 2ut' - 2ut + at^2 = 0$$

Is that correct? Then once I have the constants I can simply substitute them for $a$ $b$ and $c$ in the quadratic formula to solve for $t'$?

(Background: my question comes from my physics question here: https://physics.stackexchange.com/questions/87490/how-long-would-it-take-a-projectile-accelerating-twice-as-fast-as-another-to-cat, but I now want to solve for any, positive, acceleration of the second projectile).

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markmnl
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  • If you know the values of $u, t, a$ and $a^\prime$ then you are set and you can use the quadratic formula to solve for $t^\prime$. It looks however that you made a mistake, it should be $2a^\prime t^{\prime 2} + 2u t^\prime-2ut-at^2=0$. – ZanCoul Nov 23 '13 at 06:56
  • thanks! oops it was my original equation that was wrong, updated. – markmnl Nov 23 '13 at 10:12

1 Answers1

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Yes, this is a quadratic equation and can be solved with the quadratic formula.

$$a = a', b = 2u, c = 2ut + at^2$$

markmnl
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