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Neglecting air resistance, the distance (d) that an object fallsvaries directly as the square of the time (t) it has been falling.If an object falls 64 feet in 2 seconds, determine the distanceit will fall in 6 seconds.

Sagot :

[tex]d(t)=at^2[/tex]

where a is a constant, we can find a with the first statement:

"an object falls 64 feet in 2 seconds"

so:

[tex]\begin{gathered} 64=a(2)^2 \\ a=\frac{64}{2^2} \\ a=16 \end{gathered}[/tex]

the complete equation is

[tex]d(t)=16t^2[/tex]

now "determine the distance for 6seconds"

so, replace t=6 and find d

[tex]\begin{gathered} d(t)=16(6)^2 \\ d(t)=576ft \end{gathered}[/tex]

the distance is 576ft

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