Question

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.

Answer 1

Answer:

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)

$\lim_{t \to \infty} v = \lim_{t \to \infty} (\frac{mg}{c}[1 - e^{-\frac{ct}{m} } ] )$

$\lim_{t \to \infty} v =$ mg/c(1 - e^(-c(∞)/m))

$\lim_{t \to \infty} v =$ mg/c(1 - e^(-∞/m))

$\lim_{t \to \infty} v =$ mg/c(1 - e^(-∞))

$\lim_{t \to \infty} v =$ mg/c(1 - 0)

$\lim_{t \to \infty} v =$ mg/c(1)

$\lim_{t \to \infty} v =$ 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.