Question

A spherical ball has a diameter of 8 inches. The ball has a small puncture where air is escaping at a rate of 2.5 inches cubic per second. How long will it take for the ball to deflate? round to the nearest tenth.

Answer 1

first off, let's check how much air is in the ball, we know it has a diameter of 8, that means its radius is half that, or 4.

`\textit{volume of a sphere}\\\\ V=\cfrac{4\pi r^3}{3}~~ \begin{cases} r=radius\\[-0.5em] \hrulefill\\ r=4 \end{cases}\implies V=\cfrac{4\pi (4)^3}{3}\implies V=\cfrac{256\pi }{3}~in^3`

now, we know the air is escaping at 2.5 in³ every second, so hmm how many times does 2.5 go into V?

`\cfrac{256\pi }{3}/ 2.5\implies \cfrac{256\pi }{3}/ \cfrac{25}{10}\implies \cfrac{256\pi }{3}\cdot \cfrac{10}{25}\implies \cfrac{512\pi }{15} ~~ \approx ~~ \boxed{107.2~seconds}`