Question

The distance that an object falls from rest, when air resistance is negligible, varies directly as the square of the time that it falls (before it hits the ground). A stone dropped from rest travels 212 feet in the first 5 seconds. How far will it have fallen at the end of 7 seconds? Round your answer to the nearest integer if necessary.

Answer 1

By definition, Direct variation equations have the following form:

`y=kx`

Where "k" is the Constant of variation.

In this case let be "d" the distance in feet and "t" the time.

Since the distance that an object falls from rest, when air resistance is negligible, varies directly as the square of the time that it falls, you can write the following equation:

`d=kt^2`

You know that the stone dropped from rest and travels 212 feet in the first 5 seconds. Then you can identify that:

`\begin{gathered} d=212 \\ t=5 \end{gathered}`

Substituting values into the equation and solving for "k", you get:

`\begin{gathered} 212=k(5)^2 \\ (212)/(25)=k \\ k=8.48 \end{gathered}`

Substitute now the value of "k" and the time 7 seconds (t=7) into the equation and evaluate. Then, you get this result (rounded to the nearest integer):

`\begin{gathered} d=(8.48)(7)^2 \\ d=415.52 \\ d\approx416 \end{gathered}`

Therefore, it will have fallen 416 feet at the end of 7 seconds.