Originally Posted by

**dark_knight_307** [snip]

And here's another problem....

An object falling near the earth's surface encounters air resistance that is proportional to its velocity. The acceleration due to gravity is -9.8m/s^2. So, without air resistance the object's acceleration can be modeled by the differential equation. dv/dt = -9.8. But aerodynamic drag represents a considerable retarding force as velocity increases. Thus a better model for an object falling near the surface of the earth is: dv/dt = kv - 9.8, where k is a constant of proportionality.

1. What are the units of k? Is k positive or negative? Mr F says: Units on left have to match units on right. So kv needs to be m/s^2. And you know v is m/s .... Downwards is -ve, upwards is +ve. Gravity is down and v is down. But kv is in oposite direction to gravity ......

2. Find the velocity of the object as a function of time if the initial velocity is V sub 0. Mr F says: Solve the DE: $\displaystyle \frac{dt}{dv} = \frac{1}{kv - 9.8}$ .......

3. Use #2 to find the limit of the velocity as t -> infinity. Mr F says: Take the limit in your answer. Check: Is it the same as you get by solving 0 = kv - 9.8 (got by putting dv/dt = 0, that is, acceleration = 0 when at terminal velocity ......

4. Find the position function s(t) of the object. Mr F says: Once you have v = v(t), solve dx/dt = v(t) ......