You ride your bike to campus a distance of 55 miles and return home on the same route. Going to​ campus, you ride mostly downhill and average 99 miles per hour faster than on your return trip home. If the round trip takesnbsp one hour and nbsp one hour and 1010 minuteslong dash—that ​is, seven sixths
7
6 hourslong dash—what is your average velocity on the return​ trip? Use the formula Time traveled equals StartFraction Distance traveled Over Average velocity EndFractionTime traveled=
Distance traveled
Average velocity.
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Part 1
The average rate on the return trip is enter your response here miles per hour.

Let's break down the problem step by step and solve for the average velocity on the return trip.

### Step 1: Define variables
Let:
- \( v_r \) = the average velocity on the return trip (miles per hour).
- \( v_c = v_r + 99 \) = the average velocity on the trip to campus (99 mph faster than the return trip).

### Step 2: Use the formula for time
We are given that the round trip took 7/6 hours, and the distance each way is 55 miles. The time for each leg of the journey can be written as:

\[
t_{\text{campus}} = \frac{55}{v_c} = \frac{55}{v_r + 99}
\]

\[
t_{\text{return}} = \frac{55}{v_r}
\]

The total time for both trips is the sum of these two times, and it equals 7/6 hours:

\[
t_{\text{campus}} + t_{\text{return}} = \frac{7}{6}
\]

Substitute the expressions for the times:

\[
\frac{55}{v_r + 99} + \frac{55}{v_r} = \frac{7}{6}
\]

### Step 3: Solve for \( v_r \)
To solve this equation, we first multiply the entire equation by 6 to eliminate the fraction on the right-hand side:

\[
6 \left( \frac{55}{v_r + 99} + \frac{55}{v_r} \right) = 7
\]

Now distribute the 6:

\[
\frac{330}{v_r + 99} + \frac{330}{v_r} = 7
\]

Next, find a common denominator for the left-hand side:

\[
\frac{330 v_r + 330(v_r + 99)}{v_r(v_r + 99)} = 7
\]

Simplify the numerator:

\[
\frac{330(v_r + v_r + 99)}{v_r(v_r + 99)} = 7
\]

\[
\frac{330(2v_r + 99)}{v_r(v_r + 99)} = 7
\]

Multiply both sides by \( v_r(v_r + 99) \) to clear the denominator:

\[
330(2v_r + 99) = 7v_r(v_r + 99)
\]

Expand both sides:

\[
660v_r + 32670 = 7v_r^2 + 693v_r
\]

### Step 4: Rearrange into a quadratic equation
Move all terms to one side:

\[
7v_r^2 + 693v_r - 660v_r - 32670 = 0
\]

Simplify:

\[
7v_r^2 + 33v_r - 32670 = 0
\]

### Step 5: Solve the quadratic equation
We can now solve this quadratic equation using the quadratic formula:

\[
v_r = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}
\]

where \( a = 7 \), \( b = 33 \), and \( c = -32670 \).

Substitute these values into the quadratic formula:

\[
v_r = \frac{-33 \pm \sqrt{33^2 - 4(7)(-32670)}}{2(7)}
\]

\[
v_r = \frac{-33 \pm \sqrt{1089 + 915960}}{14}
\]

\[
v_r = \frac{-33 \pm \sqrt{917049}}{14}
\]

\[
v_r = \frac{-33 \pm 957.64}{14}
\]

### Step 6: Calculate the two possible solutions
Now solve for \( v_r \):

\[
v_r = \frac{-33 + 957.64}{14} \approx \frac{924.64}{14} \approx 66.05 \text{ mph}
\]

or

\[
v_r = \frac{-33 - 957.64}{14} \approx \frac{-990.64}{14} \approx -70.76 \text{ mph}
\]

Since velocity cannot be negative, we discard the second solution. Therefore, the average velocity on the return trip is approximately:

\[
\boxed{66.05} \text{ miles per hour}
\]

Would you like to see further details on how the quadratic formula was applied, or do you have any other questions?

Here are 5 related questions:
1. How would the solution change if the round trip time was different?
2. What is the significance of rejecting negative velocity in this context?
3. How would you find the average velocity if the return trip speed was not 99 mph slower?
4. How could we solve this using another method, such as substitution?
5. How can we interpret the result in terms of real-world biking speeds?

**Tip**: When dealing with average speeds over a round trip, remember that the speeds to and from a location impact the overall average differently due to the time spent at each speed!

You ride your bike to campus a distance of 55 miles and return home on the same route. Going to campus, you ride mostly downhill and average 99 miles per hour faster than on your return trip home. If the round trip takes 1 hour and 10 minutes—that is, seven sixths (7/6) hours—what is your average velocity on the return trip? Use the formula Time traveled = Distance traveled / Average velocity.

This problem involves calculating the average velocity on a return bike trip using algebra and the quadratic formula. The trip covers 55 miles each way, with a round trip taking 7/6 hours. Learn how to apply the quadratic formula to solve for the return trip's velocity.

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