Let 'd' represent the distance.
Let 't' represent the time.
Given that:
[tex]d\propto t^2[/tex]
Introducing a constant 'k'
[tex]d=kt^2[/tex]
where,
[tex]\begin{gathered} d=91ft \\ t=2seconds \end{gathered}[/tex]
Therefore,
[tex]\begin{gathered} 91=k\times2^2 \\ 91=k\times4 \\ Divide\text{ both sides by 4} \\ \frac{91}{4}=\frac{k\times4}{4} \\ \therefore k=\frac{91}{4}=22.75 \end{gathered}[/tex]
Hence, the relationship connecting the distance and the time is,
[tex]d=22.75t^2[/tex]
Let us now solve for the distance if the time is 6seconds.
[tex]\begin{gathered} d=22.75\times6^2=819 \\ \therefore d=819feet \end{gathered}[/tex]
Hence, the answer is 819 feet.
