Question

Keenan will run 2.5 miles from his house to Jared’s house. He plans to hang out for 45 minutes before walking home. If he can run at 6 mph and walk 4 mph, how long will he be from home?

Answer 1

Answer:

Kennan will be from home approximately an hour and 48 minutes.

Step-by-step explanation:

We must know that total time ($t_{T}$) that Keenan will be from home is the sum of run ($t_{R}$), hang out ($t_{H}$) and walk times ($t_{W}$), measured in hours:

$t_{T} = t_{R}+t_{H}+t_{W}$

If Keenan runs and walks at constant speed, then equation above can be expanded:

$t_{T} = \frac{x_{R}}{v_{R}}+t_{H}+ \frac{x_{W}}{v_{W}}$

Where:

$x_{R}$, $x_{W}$ - Run and walk distances, measured in miles.

$v_{R}$, $v_{W}$ - Run and walk speeds, measured in miles per hour.

Given that $x_{R}=x_{W} = 2.5\,mi$, $v_{R} = 6\,\frac{mi}{h}$, $v_{W} = 4\,\frac{mi}{h}$ and $t_{H} = 0.75\,h$, the total time is:

$t_{T} = \frac{2.5\,mi}{6\,\frac{mi}{h} } + 0.75\,h+\frac{2.5\,mi}{4\,\frac{mi}{h} }$

$t_{T} = 1.792\,h$ ($1\,h\,48\,m$)

Kennan will be from home approximately an hour and 48 minutes.