Question

Quinn leaves her house at 10:00 am and runs 3 miles at a steady pace of 6 miles per

Answer 1

Using the relation between velocity, distance and time, it is found that Quinn returns home at 1 pm.

Velocity is distance divided by time, that is:

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

• She walked 3 miles at a velocity of 6 miles per hour, thus $d = 3, v = 6$. The time is:

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

$t = \frac{d}{v} = \frac{3}{6} = 0.5$

• Thus, she arrived at 10:30 am, and took a lunch break of 1 hour, so she started to return at 11:30 am.
• Distance of 3 miles, velocity of 2 miles per hour, thus $d = 3, v = 2$. The time she took to return, in hours, is:

$t = \frac{d}{v} = \frac{3}{2} = 1.5$

1.5 hours after 11:30 am is 1 pm, thus, Quinn returns home at 1 pm.

A similar problem is given at brainly.com/question/24316569