Let's define the variables for the rates and times of the two parts of the trip:

• Let $r_1$ be the rate during the first part of the trip (in miles per hour).
• Let $r_2$ be the rate during the second part of the trip (in miles per hour).
• Let $t_1$ be the time for the first part of the trip (in hours).
• Let $t_2$ be the time for the second part of the trip (in hours).

We know the following information:

1. The distance for the first part of the trip is 200 miles, and for the second part, it is 225 miles.
2. The time for the second part, $t_2$, is 1 hour longer than the time for the first part, $t_1$. Hence, $t_2 = t_1 + 1$.
3. The rate during the first part, $r_1$, is 10 miles per hour faster than the rate during the second part, $r_1 = r_2 + 10$.

Now, let's use the formula $\text{Distance} = \text{Rate} \times \text{Time}$ for both parts of the trip:

For the first part: $t_1 = \frac{200}{r_1}$ For the second part: $t_2 = \frac{225}{r_2}$

From the information $t_2 = t_1 + 1$, we can substitute the expressions for $t_1$ and $t_2$:

$\frac{225}{r_2} = \frac{200}{r_1} + 1$

Also, from $r_1 = r_2 + 10$, we can substitute $r_1$ in the equation:

$\frac{225}{r_2} = \frac{200}{r_2 + 10} + 1$

Now let's solve this equation. First, multiply both sides by $r_2(r_2 + 10)$ to clear the denominators:

$225(r_2 + 10) = 200r_2 + r_2(r_2 + 10)$

Expand both sides:

$225r_2 + 2250 = 200r_2 + r_2^2 + 10r_2$

Simplify the equation:

$225r_2 + 2250 = 200r_2 + r_2^2 + 10r_2$

$225r_2 + 2250 = 210r_2 + r_2^2$

Now, move all terms to one side:

$225r_2 - 210r_2 + 2250 = r_2^2$

$15r_2 + 2250 = r_2^2$

Rearrange the equation into standard quadratic form:

$r_2^2 - 15r_2 - 2250 = 0$

Now, solve the quadratic equation using the quadratic formula. The quadratic formula is given by:

$r_2 = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$

For the equation $r_2^2 - 15r_2 - 2250 = 0$, $a = 1$, $b = -15$, and $c = -2250$. Substituting these values into the quadratic formula:

$r_2 = \frac{-(-15) \pm \sqrt{(-15)^2 - 4(1)(-2250)}}{2(1)}$

$r_2 = \frac{15 \pm \sqrt{225 + 9000}}{2}$

$r_2 = \frac{15 \pm \sqrt{9225}}{2}$

$r_2 = \frac{15 \pm 96}{2}$

So, the two possible solutions are:

$r_2 = \frac{15 + 96}{2} = \frac{111}{2} = 55.5 \quad \text{or} \quad r_2 = \frac{15 - 96}{2} = \frac{-81}{2} = -40.5$

Since speed cannot be negative, we take $r_2 = 55.5$ miles per hour.

Now, since $r_1 = r_2 + 10$, we have:

$r_1 = 55.5 + 10 = 65.5 \quad \text{miles per hour}$

Calculating the Total Time:

Now that we know the rates, we can calculate the times for each part of the trip.

For the first part: $t_1 = \frac{200}{65.5} \approx 3.05 \text{ hours}$

For the second part: $t_2 = \frac{225}{55.5} \approx 4.05 \text{ hours}$

Thus, the total time of the trip is:

$t_1 + t_2 \approx 3.05 + 4.05 = 7.1 \text{ hours}$

Final Answer:

The total time for the trip is approximately 7.1 hours.

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Would you like further explanation or a breakdown of any steps? Here are some related questions you might find helpful:

1. How can you verify the solution to this problem?
2. What happens if the rates for both parts are equal?
3. How can the relationship between time and speed be used to solve similar problems?
4. How do you apply the quadratic formula in real-world scenarios?
5. What would happen if the second part took two hours longer than the first?

Tip: Always check for the reasonableness of your solutions, especially when dealing with physical quantities like speed and time!