QuestionYou are taking a road trip in a car without A/C. The temperture in the car is 105 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 25 minutes. Given that $$\frac{T-A}{T_{0}-A}=e^{-k t}$$ where $T=$ the temperature of the pop at time $t$. $T_{0}=$ the initial temperature of the pop. $A=$ the temperature in the car. $k=$ a constant that corresponds to the warming rate. and $t=$ the length of time that the pop has been warming up.
How long will it take the pop to reach a temperature of 75 degrees $F$ ? It will take $\square$ minutes.

Studdy Solution
Use the value of $k$ to find the time $t$ when the pop reaches 75 degrees F.
Substitute $T = 75$ into the original formula:
$\frac{75 - 105}{45 - 105} = e^{-kt}$
Calculate the left side of the equation:
$\frac{-30}{-60} = \frac{1}{2}$
Thus, the equation becomes:
$\frac{1}{2} = e^{-kt}$
Substitute the expression for $k$ from STEP_2:
$\frac{1}{2} = e^{\left(\frac{\ln\left(\frac{3}{4}\right)}{25}\right)t}$
Take the natural logarithm of both sides:
$\ln\left(\frac{1}{2}\right) = \left(\frac{\ln\left(\frac{3}{4}\right)}{25}\right)t$
Solve for $t$:
$t = \frac{\ln\left(\frac{1}{2}\right)}{\frac{\ln\left(\frac{3}{4}\right)}{25}}$
Calculate $t$:
$t \approx 57.78$
Therefore, it will take approximately $\boxed{58}$ minutes for the pop to reach 75 degrees F.