The problem states:

• Your friend averages 56 miles per hour while driving to a college.
• On the return trip, your friend averages 48 miles per hour due to construction.
• The total driving time for both trips is 6 hours and 30 minutes (which is 6.5 hours).
• We need to determine how long each trip takes.

Let:

• $t_1$ be the time for the trip to college (at 56 mph).
• $t_2$ be the time for the return trip (at 48 mph).
• The total time for both trips is given by $t_1 + t_2 = 6.5$ hours.

Step 1: Express the distances for both trips

Since the distance to and from the college is the same, we can set up the following distance equations for both legs of the trip.

The distance to the college is: $\text{Distance} = 56 \cdot t_1$ The distance on the return trip is: $\text{Distance} = 48 \cdot t_2$ Since these distances are the same: $56 \cdot t_1 = 48 \cdot t_2$ This gives us the first equation: $t_2 = \frac{56}{48} \cdot t_1 = \frac{7}{6} \cdot t_1$

Step 2: Use the total time equation

We also know that the total time for both trips is 6.5 hours: $t_1 + t_2 = 6.5$ Substitute $t_2 = \frac{7}{6} t_1$ into this equation: $t_1 + \frac{7}{6} t_1 = 6.5$ $\left(1 + \frac{7}{6}\right) t_1 = 6.5$ $\frac{13}{6} t_1 = 6.5$ Now solve for $t_1$: $t_1 = \frac{6.5 \times 6}{13} = 3 \text{ hours}$

Step 3: Solve for $t_2$

Now that we know $t_1 = 3$, we can find $t_2$: $t_2 = \frac{7}{6} \times 3 = 3.5 \text{ hours}$

Conclusion:

• The trip to college takes 3 hours.
• The return trip takes 3.5 hours.

Would you like further details or have any questions about this solution?

Here are some related questions to deepen your understanding:

1. How would the result change if the speeds were switched?
2. How can you calculate the total distance traveled using this information?
3. What would happen if there was a different total time?
4. How can this problem be solved graphically using distance vs. time graphs?
5. How can you calculate average speed for the entire journey?

Tip: Always double-check your units when working with time and speed to avoid calculation errors.