Starting at home, ishaan traveled uphill to the grocery store for 121212 minutes at just 151515 mph. he then traveled back home along the same path downhill at a speed of 303030 mph. what is his average speed for the entire trip from home to the grocery store and back?

The average speed is defined as the total distance covered divided by the time interval.

First of all we need to convert the time interval from minutes to hours, so:

[tex]121212min(\frac{1h}{60min}) = 2020.2h[/tex]

We know that the distance, speed and time are related as follows:

[tex]d = vt[/tex]

Thus, the distance covered from home to the grocery store is:

[tex]d = 151515(2020.2)=306090603mi[/tex] 

So, the total distance covered for the entire trip from home to the grocery store and back is:

[tex]d_{t} = 2(306090603) = 612181206mi[/tex] 

We need to the total time interval, that is:

[tex]t = t_{1} + t_{2} = 2020.2h + t_{2}[/tex]

So, it is necessary to find [tex]t_{2} [/tex] as:

[tex]t_{2} = \frac{306090603mi}{303030mph} = 1010.1h[/tex]

In this way:

[tex]t = 2020.2h + 1010.1h = 3030.3h [/tex]

Finally, his average speed for the entire trip from home to the grocery store and back is:

[tex]v_{a} = \frac{d_{t}}{t} = \frac{612181206mi}{3030.3h} = 202020,ph[/tex]