Persons on a hiking trip move with different average speed on different days. The percentage difference of the speed between these two days 2.5%.
Given:
On the second day of a hiking trip, the tourists covered 17% more distance than on the first day. On the second day they were hiking for 20% longer time.
Let d be the distance covered on first day and t be the time spent on the first day.
So, the distance covered on second day will be,
[tex]d'=d+17\%\rm \;of \; \it d\\d'=\rm 1.17\it d[/tex]
Similarly, the time spent on second day will be,
[tex]t'=t+20\%\rm \;of \; \it t\\t'=\rm 1.20\it t[/tex]
Now, the speed on first day is [tex]s=\dfrac{d}{t}[/tex].
The speed on second day will be,
[tex]s'=\dfrac{d'}{t'}\\s'=\dfrac{1.17d}{1.20t}\\s'=0.975\dfrac{d}{t}\\s'=0.975s[/tex]
So, the percentage difference of the speed between these two days will be,
[tex](s-s')\times 100\%=(s-0.975s)\times 100\%\\=0.025\times 100\%\\=2.5\%[/tex]
Therefore, the percentage difference of the speed between these two days 2.5%.
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https://brainly.com/question/21807125
