# How is gravity a quadratic function?

Jul 13, 2015

Refer to explanation.

#### Explanation:

We can apply quadratic functions to objects that are in motion under gravity. For this explanation, we take a look at one of the equations of motion from physics that itself is a quadratic function and we set the acceleration of an object as being influenced by gravity.

Recall that a quadratic equation looks like the following:

$f \left(x\right) = a {x}^{2} + b x + c$

If we were to recall one of equations of motion from physics, such as the equation:

${x}_{f} - {x}_{0} = {v}_{0} \cdot t + \frac{1}{2} a \cdot {t}^{2}$

where, ${x}_{f} - {x}_{0} =$change in position of $x$$= \Delta x$

and if we consider our acceleration, $a$, to be the gravitational acceleration $g$, then

$\Delta x = {v}_{0} \cdot t + \frac{1}{2} g \cdot {t}^{2}$

rearranging this equation gives

$\Delta x = \frac{1}{2} g \cdot {t}^{2} + {v}_{0} \cdot t$

and making the change in the position of $x$ with respect to time $t$ gives

$\Delta x \left(t\right) = \frac{1}{2} g \cdot {t}^{2} + {v}_{0} \cdot t$

the $\frac{1}{2} g$ part is like the $a$ part in $f \left(x\right) = a {x}^{2} + b x + c$

the ${v}_{0}$ part is like the $b$ part in $f \left(x\right) = a {x}^{2} + b x + c$

$c$ is just some constant (some number), so think of $c$ in the equation $\Delta x \left(t\right) = \frac{1}{2} a \cdot {t}^{2} + {v}_{0} \cdot t$
as being equal to $0$, $\left(c = 0\right)$

the $t$'s are like the $x$'s in $f \left(x\right) = a {x}^{2} + b x + c$

So, if we look at them together:

$f \left(x\right) = \left(a\right) {x}^{2} + \left(b\right) x + c$

$\Delta x \left(t\right) = \left(\frac{1}{2} a\right) {t}^{2} + \left({v}_{0}\right) t$