Newton’s law of cooling states that dx/dt= −k(x − A) where x is the temperature, t is time, A is the ambient temperature, and k > 0 is a constant. Suppose that A A0 cos(ωt) for some constants A0 and ω. That is, the ambient temperature oscillates (for example night and day temperatures).a. Find the general solution. b. In the long term, will the initial conditions make much of a difference? Why or why not?

(a) The solution to the differential equation is x = A_0Coswt + Ce^(-kx)

(b) The initial condition t > 0 will not make much of a difference.

Step-by-step explanation:

Given the differential equation

dx/dt= −k(x − A); t > 0, A = A_0Coswt

(a) To solve the differential equation, first separate the variables.

dx/(x - A) = -kdt

Integrating both sides, we have

ln(x - A) = -kt + c

x - A = Ce^(-kt) (Where C = ce^(-kt))

x = A + Ce^(-kx)

Now, we put A = A_0Coswt

x = A_0Coswt + Ce^(-kx) (Where C is constant.)

And we have the solution.

(b) Since temperature t ≠ 0, the initial condition t > 0 will not make much of a difference because, Cos(wt) = Cos(-wt).

It is not any different from when t < 0.

sine

step-by-step explanation:

54 kilometers per hour

step-by-step explanation:

according to rules:

1 km = 1000 meter

which means that 1 meter = 1/1000 km

1 hour = 3600 second [ as 1 hour = 60 min = 60 *60 seconds ]

which means that 1 second = 1/3600 hour

therefore, 1 kilometer/hour =

thus to convert 15 meter per second to kilometer per hour, we multiply 15 to 3.6 kilometer/hour.

15 meter/second=   which is the required answer.