A playground swing sways back and forth and loses momentum and distance with each oscillation. If the swing travels a total distance of 8 feet on the first iteration and the distance of each consecutive oscillation is 3/4 of the previous distance, what distance will the swing travel in the third oscillation, measured to the nearest whole foot?

Answer 1

To find the distance traveled by the swing in the third oscillation, we first note that the distance in each oscillation reduces to 3/4 of the distance of the previous oscillation.

Let's denote the distance traveled in the first oscillation as \( d_1 = 8 \) feet.

For the second oscillation, the distance is given by:

\[ d_2 = \frac{3}{4} d_1 = \frac{3}{4} \times 8 = 6 \text{ feet} \]

For the third oscillation, the distance is:

\[ d_3 = \frac{3}{4} d_2 = \frac{3}{4} \times 6 \]

Calculating this:

\[ d_3 = \frac{3 \times 6}{4} = \frac{18}{4} = 4.5 \text{ feet} \]

Rounding \( 4.5 \) to the nearest whole foot, we get:

\[ d_3 \approx 5 \text{ feet} \]

Thus, the distance the swing will travel in the third oscillation, measured to the nearest whole foot, is \(\boxed{5}\).