Answer:
50 minutes
Step-by-step explanation:
Given:
Temperature, T = 230°F
Overheat temp, To = 280°F
Ts = 80°F
Cooling rate, r = 0.0058
Since the car can be driven again at 230°F, and when it is 80°F outside, the cooling rate of the car is 0.0058, using Newton's law of cooling, we have:
[tex] T = (T_o - T_s)​ e^-^r^t + T_s[/tex]
Substituting figures, we have:
[tex] 230 = (280 - 80)​ e^-^0^.^0^0^5^8^t + 80[/tex]
[tex] 230 = 200e^-^0^.^0^0^5^8^t + 80[/tex]
Solving further, let's subtract 80 from both sides, we now have:
[tex] 230 - 80= 200e^-^0^.^0^0^5^8^t + 80 - 80[/tex]
[tex] 150 = 200e^-^0^.^0^0^5^8^t [/tex]
Divide both sides by 200:
[tex] \frac{150}{200}= \frac{200e^-^0^.^0^0^5^8^t}{200}[/tex]
[tex] = \frac{150}{200}=e^-^0^.^0^0^5^8^t[/tex]
Let's take the natural log of both sides, we have:
[tex] In 0.75 = In e^-^0^.^0^0^5^8^t[/tex]
Using logarithm power rule, we have:
= - 0.2877 = - 0.0058t
To find t, let's now divide both sides by -0.0058
[tex] \frac{-0.2877}{-0.0058} = \frac{-0.0058t}{-0.0058} [/tex]
50 = t
Therefore, you have to wait for 50 minutes before you can continue driving.
