Given that:
A bee flies 10 ft/s directly towards a flowerbed from its hive.
It stays at the flowerbed for 15 minutes and then flies directly back to the hive at 8 ft/s.
The total time it is away from there is 19 minutes.
Consider that speed is the ratio of distance traveled to the time taken. That is:
[tex]s=\frac{d}{t}[/tex]Let x represent the distance from the bee hive to the flowerbed, and d be the distance.
When speed was 10 ft/s, let the time be y
[tex]\begin{gathered} 10=\frac{d}{y_{}} \\ \\ d=10y \end{gathered}[/tex]When the speed was 8 ft/s, let time be z
[tex]\begin{gathered} 8=\frac{d}{z} \\ \\ d=8z \end{gathered}[/tex]Distance is unchanged, so
[tex]10y=8z\ldots\ldots\ldots\ldots\ldots\ldots\ldots\ldots\ldots\ldots\ldots\ldots\ldots\ldots\ldots\text{.}(1)[/tex]Since the total time is 19 minutes = 1140 seconds
We have the equation for time as:
[tex]y+z=1140\ldots\ldots\ldots\ldots\ldots\ldots\ldots\ldots\ldots\ldots\ldots\ldots\ldots\ldots\ldots\ldots\ldots\text{.}\mathrm{}(2)[/tex]Solving (1) and (2) simultaneously, we can obtain the values for x and y, and these will help know the value for the required distance.