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This problem involves Newton's Law of Cooling (or warming in this case), modeled by the equation:

\[
\frac{T - A}{T_0 - A} = e^{-kt}
\]

Where:
- \(T\) is the temperature of the pop at time \(t\),
- \(T_0 = 45^\circ F\) is the initial temperature of the pop,
- \(A = 85^\circ F\) is the ambient temperature (temperature in the car),
- \(t\) is the time in minutes,
- \(k\) is a constant that characterizes the rate of warming.

### Step 1: Solve for the constant \(k\)
We are given that the pop's temperature reaches 60°F after 10 minutes. So, we can plug these values into the formula to find \(k\).

For \(T = 60^\circ F\) and \(t = 10\):

\[
\frac{60 - 85}{45 - 85} = e^{-10k}
\]

This simplifies to:

\[
\frac{-25}{-40} = e^{-10k}
\]

\[
\frac{5}{8} = e^{-10k}
\]

Take the natural logarithm of both sides:

\[
\ln\left(\frac{5}{8}\right) = -10k
\]

\[
k = -\frac{1}{10} \ln\left(\frac{5}{8}\right)
\]

Now, calculate \(k\):

\[
k \approx -\frac{1}{10} \cdot (-0.470) = 0.047
\]

### Step 2: Find the time for the pop to reach 74.75°F
Now, use the same equation to find the time \(t\) when \(T = 74.75^\circ F\).

\[
\frac{74.75 - 85}{45 - 85} = e^{-kt}
\]

Simplifies to:

\[
\frac{-10.25}{-40} = e^{-0.047t}
\]

\[
0.25625 = e^{-0.047t}
\]

Take the natural logarithm of both sides:

\[
\ln(0.25625) = -0.047t
\]

\[
t = \frac{\ln(0.25625)}{-0.047}
\]

\[
t \approx \frac{-1.358}{-0.047} \approx 28.89 \text{ minutes}
\]

Thus, it will take approximately **28.89 minutes** for the pop to reach a temperature of 74.75°F.

Would you like more details or further clarification?

Here are some related questions for further understanding:
1. How does the value of \(k\) affect the warming rate of the pop?
2. What would happen to the time \(t\) if the initial temperature \(T_0\) were lower?
3. How does the formula change if the ambient temperature \(A\) decreases?
4. Can you derive the time it takes for the pop to reach 85°F?
5. How would the process differ if cooling instead of warming was modeled?

**Tip:** Newton's Law of Cooling can be applied to both heating and cooling processes, depending on the context of the temperature difference between the object and its surroundings.

You are taking a road trip in a car without A/C. The temperature in the car is 85 degrees F. You buy a cold pop at a gas station. Its initial temperature is 45 degrees F. The pop's temperature reaches 60 degrees F after 10 minutes. Given that (T - A) / (T0 - A) = e^(-kt), how long will it take the pop to reach a temperature of 74.75 degrees F?

Solve a real-world problem using Newton's Law of Cooling. This question explores exponential equations by modeling how a cold drink warms in a car at 85°F. Learn how to determine the constant warming rate and calculate the time needed for the drink to reach 74.75°F.

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