How would I rearrange the equation $S = ut + \frac 12 at^2$ to make $t$ the subject of the equation?

Question

How would I rearrange the equation $S = ut + \frac 12 at^2$ to make t the subject of the equation ?

I'm not sure whether I should times the left side by $2$ or minus $\frac 12$ from the left side.

Answer 1

Let $s=ut+\frac{1}{2}at^2$

Re-write \begin{align} t^{2}+\frac{2u}{a}t-\frac{2s}{a}=0 \end{align} Thus, \begin{align} t &= \frac{-\frac{2u}{a} \pm \sqrt{\frac{4u^{2}}{a^{2}}+\frac{8s}{a}}}{2}\\ &= -\frac{u}{a} \pm \sqrt{\frac{u^{2}}{a^{2}}+\frac{2s}{a}} \end{align}

Answer 2

This is a quadratic in $t$. Meaning you should use the quadratic formula.

Moving the "$S$" constant to the right, we get $\frac 12 at^2+ut-S=0$ And multiplying by $2$, we get$$at^2+2ut-2S=0$$

Solving for $t$, we get $$t=\frac {-2u\pm\sqrt{4u^2-4(-2S)(a)}}{2a}=\frac {-2u\pm\sqrt{4u^2+8aS}}{2a}=\frac {-u\pm\sqrt{u^2+2aS}}{a}$$

Answer 3

The quadratic solution is obvious as shown in earlier solutions posted here. What is important to note is that in for the standard case, time is positive hence $t>0$, i.e. we can only take the positive root, i.e. $$t=\frac {-u+\sqrt{u^2+2aS}}a$$

Answer 4

To be hyperprecise, the answer is

$$t= \begin{cases} \displaystyle{-u\pm\sqrt{u^2+2aS}\over a}&\text{if }a\not=0\\ \displaystyle{S\over u}&\text{if }a=0\text{ and } u\not=0\\ \end{cases}$$

If $a=u=0$ the equation becomes $S=0$, which is either true for all $t$ or for no $t$, depending on the value of $S$.