I am trying to solve for $T$ in the following equation: $$V_f = V_i + AT + \frac 1 2JT^2.$$ Does anyone know how? I tried myself but since I'm 33 I can no longer remember basic algebra :(.


  • $\begingroup$ If as in the edit the last term is $\frac{1}{2}JT^2$, you use the Quadratic Formula. If the last term is supposed to be $\frac{1}{2JT^2}$, then in principle one could use the Cardano Formula for cubics, but in practice a numerical method is best. $\endgroup$ – André Nicolas Jul 28 '13 at 23:56
  • $\begingroup$ @AndréNicolas It looks like a physics equation $V_i$, $V_f$ are initial and final velocity, $J$ (jerk?)and $A$ (acceleration?) should be other magnitudes, so $\frac 1 2JT^2$ makes sense. $\endgroup$ – Pedro Tamaroff Jul 28 '13 at 23:57
  • $\begingroup$ You are undoubtedly right. My monitor is too fuzzy for me to read subscripts. $\endgroup$ – André Nicolas Jul 28 '13 at 23:59
  • $\begingroup$ @PeterTamaroff, it looks like you're right about jerk. $\endgroup$ – dfeuer Jul 29 '13 at 1:32
  • $\begingroup$ yes I am trying to figure out the stopping distance of a ship in water knowing only velocity and time. The deceleration rate is not constant, which would require jerk. Thanks! $\endgroup$ – wayofthefuture Jul 29 '13 at 3:09

Hint You have a quadratic equation in $T$. Recall that we can solve $$aT^2+bT+c=0$$ using the formula $$T=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$$

You have $$ (V_i-V_f) + AT + \frac 1 2JT^2=0$$

so $$\begin{align}V_i-V_f=&c\;\\A=&b\\\frac 1 2J=&a\end{align}$$

  • $\begingroup$ LOL, you are assuming way too much about my algebra skills $\endgroup$ – wayofthefuture Jul 28 '13 at 23:54
  • $\begingroup$ @scuzzlebuzzle What about now? $\endgroup$ – Pedro Tamaroff Jul 28 '13 at 23:58
  • $\begingroup$ (-A + Sqrt(A^2 - 4(1/2J)(Vi-Vf))) / 2(1/2J) ?? $\endgroup$ – wayofthefuture Jul 29 '13 at 1:51
  • $\begingroup$ If you mean $$\frac{{ - A \pm \sqrt {{A^2} - 2J\left( {{V_i} - {V_f}} \right)} }}{J}$$ then yes. $\endgroup$ – Pedro Tamaroff Jul 29 '13 at 2:04
  • $\begingroup$ Thank You Peter. I know it takes time to write that stuff out, I really appreciate it. $\endgroup$ – wayofthefuture Jul 29 '13 at 2:21

In general, given a quadratic equation $$x^2+px+q=0,$$ we can "complete the square" to turn this into something we can solve more easily.

Specifically, we first rewrite the equation as $$x^2+2\left(\frac12 p\right)x+q=0.$$

Note, then that because $(x+w)^2=x^2+2w+w^2$, we can write $$\left(x+\frac 1 2 p\right)^2-\left(\frac 1 2 p\right)^2+q=0.$$ Then $$x+\frac12 p=\pm\sqrt {\frac{p^2}4-q},$$ so $$x=\frac{-p\pm\sqrt{p^2-4q}}{2}.$$ Now in general, $x^2$ may have a coefficient other than $1$. If we have an equation $$ax^2+bx+c=0,$$ we divide through by $a$ to get $$x^2+\frac b a x + \frac c a = 0,$$ so $$\begin{array}{cc}p=\frac b a & q=\frac c a\end{array}$$ and substituting gets us $$x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}.$$


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