Ive been given the 4 kinematic equations for constant acceleration. The fourth being:
$$s=ut+\frac{1}{2}at^2.$$
If rearranged it forms the quadratic equation
$$at^2+2ut-2s=0.$$
But that means that $t$ has 2 values.
Will one of them always be negative? So only one value is realistically possible?
And how could you rearrange it to get $t$ on it's own? Would that be using the quadratic formula?
Answer
The equation describes parabolic motion, if $a\neq 0$ is a non-zero constant acceleration, which I will assume from now on. If you think about it, your solution provides an answer to the question: at what time does the object is in the position $s$? [A note on notation: Traditionally, the letter $s$ denotes distance (I guess from the German word "Strecke"), which by definition is a non-negative quantity, but your formula makes more sense, if we interpret $s$ as a position $x$, which can also be negative.]
$$\frac{1}{2}at^2+ut-x=0$$
$$t_1 = \frac{-u-\sqrt{D}}{a}$$
$$t_2 = \frac{-u+\sqrt{D}}{a}$$
where $D := u^2+2ax$. Let's think about it for a moment, and see what answers we get by varying $x$.
Case $D<0$:
The discriminant is negative, there are no solutions, therefore at no time your object will have that position.
Case $D=0 \Leftrightarrow x=-\frac{u^2}{2a}$:
The discriminant is zero, there is only one solution which is the "top" ("bottom") point reached by the object,if $a<0$ ($a>0$), respectively.
Case $D>0$:
The discriminant is positive and there are two solutions. This means that the object will reach that position twice, once going "up" and once going "down" the parabola.
Now, will one of them always be negative? Not necessarily.
Case $ax > 0$: One positive and one negative.
Case $ax < 0$ and $\frac{u}{a}>0$: Two negative.
Case $ax < 0$ and $\frac{u}{a}<0$: Two positive.
Case $x = 0$ and $\frac{u}{a}>0$: One zero and one negative.
Case $x = 0$ and $\frac{u}{a}<0$: One zero and one positive.
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