In the equation $x = \frac{1}{3}gt^2$, if g remains constant and x equals –9 when t is 3, what will be the value of x when t is 5?

We start with the given equation:

$x = \frac{1}{3} g t^2$

Step 1: Solve for $g$

We are given that when $t = 3$, $x = -9$. Substituting these values:

$-9 = \frac{1}{3} g (3)^2$

$-9 = \frac{1}{3} g (9)$

$-9 = 3g$

$g = -3$

Step 2: Find $x$ when $t = 5$