Question

Supongamos que el lector

está manejando su auto en un frío día de invierno (20ºF al exterior) y el motor se sobrecalienta (a unos 220ºF). Cuando se estaciona, el motor empieza a enfriarse. La temperatura T del motor
t minutos después de estacionarlo satisface la ecuación
ln a
T 20
200 b 0.11t
(a) De la ecuación, despeje T.
(b) Use la parte (a) para hallar la temperatura del motor después de 20 minutos 1t 202.

Answer 1

T20 would bet the settings for its doing t

Answer 2

a. The temperature of the motor after t minutes is: T(t) = 20 + 200e^(-0.11t)

b. The temperature of the motor after 20 minutes is approximately 89.6°F.

Part (a)

The equation is:

ln(T-20)/200=-0.11t

To solve for T, we can exponentiate both sides of the equation:

T-20 = 200e^(-0.11t)

Adding 20 to both sides, we get:

T = 20 + 200e^(-0.11t)

Therefore, the temperature of the motor after t minutes is:

T(t) = 20 + 200e^(-0.11t)

Part (b)

To find the temperature of the motor after 20 minutes, we can simply substitute t = 20 into the equation for T(t):

T(20) = 20 + 200e^(-0.11*20) ≈ 89.6°F

Therefore, the temperature of the motor after 20 minutes is approximately 89.6°F.