Question

On the second day of a hiking trip, a group of tourists walked a 17% greater distance than on the first day. By what percent did their average speed change from the first day to the second day if on the second day they hiked for 20% longer?
HELP ME PLZ.

Answer 1

Using the relation between velocity, distance and time, it is found that their average speed changed by -2.5%.

Velocity is distance divided by time, that is:

$v = \frac{d}{t}$

In this problem:

• 17% greater distance than on the first day, hence $d = 1.17d$.

• Hiked for 20% longer, hence $t = 1.2t$

Then:

$v_2 = \frac{d}{t} = \frac{1.17d}{1.2t} = \frac{1.17}{1.2}v = 0.975$

1 - 0.975 = 0.025 = 2.5%

Their average speed changed by -2.5%.

To learn more about velocity, distance and time, you can take a look at https://brainly.com/question/24316569