Falling skydiver differential equation

  • Thread starter darryw
  • Start date
  • #1
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Homework Statement


The velocity v(t) of a skydiver falling to the ground is governed by

m dv/dt = mg - kv^2

where m is the skydiver's mass, g is the acceleration due to gravity, k > 0 is the drag
coefficient, and v(t) >= 0.

Solve this equation for v(t) with the initial condition v(0) = 0.


Homework Equations



I have been doing other diff equations but when its a word problem it confuses me.. So, i treat m, g and k as constants.
I rearranged equation into this:

dv/dt + (k/m)v^2 = g

so now i can use integrating factor..

mu(x) = e^integ (k/m) ...but im going to stop here already, in case im already on wrong track.. The fact that v is squared changes things, no? (cant use integrating factor?) If this is ok though, i will proceed.. Thanks alot for any help/tips

The Attempt at a Solution


Homework Statement





Homework Equations





The Attempt at a Solution


Homework Statement





Homework Equations





The Attempt at a Solution


Homework Statement





Homework Equations





The Attempt at a Solution

 

Answers and Replies

  • #2
rock.freak667
Homework Helper
6,230
31

Homework Equations



I have been doing other diff equations but when its a word problem it confuses me.. So, i treat m, g and k as constants.
I rearranged equation into this:

dv/dt + (k/m)v^2 = g

so now i can use integrating factor..

mu(x) = e^integ (k/m) ...but im going to stop here already, in case im already on wrong track.. The fact that v is squared changes things, no? (cant use integrating factor?) If this is ok though, i will proceed.. Thanks alot for any help/tips
Instead of putting it in that form, put it in a different form

[tex]\frac{dv}{dt} = g- \frac{k}{m}v^2[/tex]

[tex]\frac{1}{g- \frac{k}{m}v^2} \frac{dv}{dt} = 1[/tex]

Split into partial fractions.
 
  • #3
127
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I dont understand the reason for putting it in that form?
 
  • #4
rock.freak667
Homework Helper
6,230
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I dont understand the reason for putting it in that form?
You had it in the form of

dv/dt + (k/m)v2 = g


this will not give you an integrating factor of e∫(k/m)dt


had the v2, just been 'v', then it would work fine.
 
  • #5
127
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ok thankjs, .so in the form you have it i can get an integ factor? I mean, this is the method i need to use to solve right? (integrating factor) (as opposed to separable equation)
 
  • #6
34,174
5,788
ok thankjs, .so in the form you have it i can get an integ factor? I mean, this is the method i need to use to solve right? (integrating factor) (as opposed to separable equation)
Forget integrating factors for this problem. The integrating factor business applies to problems in this form:
y' + P(x)y = Q(x)
What you have doesn't fit in this form.

rock.freak667 has shown you how to separate the equation and has given you advice on how to continue.
 
  • #7
127
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I need to confirm something before I proceed solving: in this equation, t in independent variable, v is dependent variable. k,m, and g are all constants. Is that correct?

Also, if i compare this to the previous equations ive been solving, t is like x (independent), and v is like y (dependent). and also t will always be independent because its time in these motion differential equations right? thanks
 
  • #8
rock.freak667
Homework Helper
6,230
31
Yes.
 

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