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Question
- the distance s that an object falls varies directly with the square of the time, t, of the fall. if an object falls 16 feet in one second, how long for it to fall 144 feet?
Step1: Set up the direct - variation equation
Since $s$ varies directly with $t^{2}$, we have $s = kt^{2}$. When $s = 16$ and $t = 1$, we substitute these values into the equation: $16=k\times1^{2}$, so $k = 16$. The equation becomes $s = 16t^{2}$.
Step2: Solve for $t$ when $s = 144$
Substitute $s = 144$ into the equation $s = 16t^{2}$: $144=16t^{2}$. First, divide both sides by 16: $\frac{144}{16}=t^{2}$, which simplifies to $9=t^{2}$. Then, take the square root of both sides. Since $t>0$ (time cannot be negative in this context), $t = 3$.
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