Answer:
Kennan will be from home approximately an hour and 48 minutes.
Step-by-step explanation:
We must know that total time (
) that Keenan will be from home is the sum of run (
), hang out (
) and walk times (
), measured in hours:
![t_{T} = t_{R}+t_{H}+t_{W}](https://tex.z-dn.net/?f=t_%7BT%7D%20%3D%20t_%7BR%7D%2Bt_%7BH%7D%2Bt_%7BW%7D)
If Keenan runs and walks at constant speed, then equation above can be expanded:
![t_{T} = \frac{x_{R}}{v_{R}}+t_{H}+ \frac{x_{W}}{v_{W}}](https://tex.z-dn.net/?f=t_%7BT%7D%20%3D%20%5Cfrac%7Bx_%7BR%7D%7D%7Bv_%7BR%7D%7D%2Bt_%7BH%7D%2B%20%5Cfrac%7Bx_%7BW%7D%7D%7Bv_%7BW%7D%7D)
Where:
,
- Run and walk distances, measured in miles.
,
- Run and walk speeds, measured in miles per hour.
Given that
,
,
and
, the total time is:
![t_{T} = \frac{2.5\,mi}{6\,\frac{mi}{h} } + 0.75\,h+\frac{2.5\,mi}{4\,\frac{mi}{h} }](https://tex.z-dn.net/?f=t_%7BT%7D%20%3D%20%5Cfrac%7B2.5%5C%2Cmi%7D%7B6%5C%2C%5Cfrac%7Bmi%7D%7Bh%7D%20%7D%20%2B%200.75%5C%2Ch%2B%5Cfrac%7B2.5%5C%2Cmi%7D%7B4%5C%2C%5Cfrac%7Bmi%7D%7Bh%7D%20%7D)
(
)
Kennan will be from home approximately an hour and 48 minutes.