Making an expression an explicit function of time

[leehufford]

Hello,
This actually isn't a homework question, more of a curiosity, but since its a book problem I decided to post it here.

Homework Statement

An object is released from rest at time t=0 and falls through the air, which exerts a resistive force such that the acceleration a of the the object is given by a = g - bv, where v is the objects speed and b is a constant. If limiting cases for large and small values of t are considered, which of the following is a possible expression for the speed of the object as an explicit function of time?

Homework Equations

This sort of looks like a differential equation, but the book I am reading this from assumes you haven't taken that class yet. Aside from the solution, I was wondering why they put the statement about limiting cases being considered. It's easy to solve for v. The correct answer is

v = g(1-e^-bt)/b.

The Attempt at a Solution

I can get v = (g-a)/b. The e^-bt looks like a complimentary solution to a homogenous constant coefficient differential equation, but again, I don't think students using this book know about DE's. This one has me stumped. Any help would be greatly appreciated. Thanks in advance,

Lee

 

[CAF123]

If you haven't covered DE's yet, then I guess they want you to analyse the functional form of v and see if it makes sense. For example, assuming the object is dropped from a great enough height, what happens to the velocity of the object at large enough t? What would you expect for small t?

Are your thoughts consistent with the provided equation for v?

 

[leehufford]

If you haven't covered DE's yet, then I guess they want you to analyse the functional form of v and see if it makes sense. For example, assuming the object is dropped from a great enough height, what happens to the velocity of the object at large enough t? What would you expect for small t?

Are your thoughts consistent with the provided equation for v?

I suppose I should've given the other possible answers, I just assumed there was a non DE way to actually solve this, rather than eliminating the other choices. The choices were:

a) v = g(1-e^-bt)/b

b) v = (ge^bt)/b

c) v = gt-bt^2

d) v = (g+a)t/b

e) v = v₀ +gt

Choice E can be eliminated because there is no b, and certainly b influences the speed. I guess my question is how would you know to arrive at an exponential function? I can honestly only eliminate E on my own.

By the way, I am taking DE now (about 2/3 of the way through) so If you would like to both explain the elimination process without DE and also explain how to derive this with DE's, that would be extremely awesome. Thanks for the reply!

-Lee

 

[CAF123]

I guess my question is how would you know to arrive at an exponential function?

I don't think intuitively there is a reason why the velocity falls off like a decaying exponential. But notice that as t gets large, this exp tends to zero. What then happens to v in the expression they gave for the solution? Does this make sense?

To decide if it makes sense try recalling what happens to the velocity of an object if it is allowed to fall for long enough. Then also try to see if the other solutions agree with this.

By the way, I am taking DE now (about 2/3 of the way through) so If you would like to both explain the elimination process without DE and also explain how to derive this with DE's, that would be extremely awesome.

Reexpress a = dv/dt. g and b are constants. Can you see how to proceed?

 

[leehufford]

I don't think intuitively there is a reason why the velocity falls off like a decaying exponential. But notice that as t gets large, this exp tends to zero. What then happens to v in the expression they gave for the solution? Does this make sense?

To decide if it makes sense try recalling what happens to the velocity of an object if it is allowed to fall for long enough. Then also try to see if the other solutions agree with this.


Reexpress a = dv/dt. g and b are constants. Can you see how to proceed?

Absolutely. A simple integrating factor for a first order ODE. I also understand the elimination process now. Thank you so much for your replies!

-Lee