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

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

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) = (x_(R))/(v_(R))+t_(H)+ (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\,(mi)/(h)`,
`v_(W) = 4\,(mi)/(h)` and
`t_(H) = 0.75\,h`, the total time is:

`t_(T) = (2.5\,mi)/(6\,(mi)/(h) ) + 0.75\,h+(2.5\,mi)/(4\,(mi)/(h) )`

`t_(T) = 1.792\,h` (
`1\,h\,48\,m`)

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