If an object of mass m is dropped from rest, one model for its speed v after t seconds, taking air resistance into account is: c= mg/c (1- e^ -ct/m), where g is the acceleration due to gravity and c is a positive constant describing air resistance. Required:
a. Calculate lim v
t→[infinity]
b. What is the meaning of this limit? (choose from the following options)

1. It is the time it takes the object to reach its maximum speed.
2. It is the speed the object reaches before it starts to slow down.
3. It is the time it takes the object to stop.
4. It is the speed the object approaches as time goes on.

a. mg/c b. 4. It is the speed the object approaches as time goes on.

Step-by-step explanation:

a. Calculate lim v  as t→[infinity]

Since v = mg/c(1 - e^ -ct/m)

mg/c(1 - e^(-c(∞)/m))

mg/c(1 - e^(-∞/m))

mg/c(1 - e^(-∞))

mg/c(1 - 0)

mg/c(1)

mg/c

b. What is the meaning of this limit?

4. It is the speed the object approaches as time goes on.

This is because, since t → ∞ implies a long time after t = 0, the limit of v as t → ∞ implies the speed of the object after a long time. So, the limit of v as t → ∞ is the speed the object approaches as time goes on.

  b.   the average rate of change of f(x) is less than the average rate of change of g(x).

step-by-step explanation:

since the length of the interval is the same in each case, we only need to compare the change in y-value to see which is greater.

so the change in y-value for function f is (6 )) = 9.

the table tells us

so the change in y-value for function g is (31 -1) = 30.

the change in y-value of f(x) on the interval is less than the change in y-value of g(x) on the same interval, so the average rate of change of f(x) is less than the average rate of change of g(x).

well what did you do it matters what you did : )

step-by-step explanation: