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Change of variable dif equation problem

by spoon
Tags: equation, variable
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spoon
#1
Jul24-05, 07:18 PM
P: 12
Driving your 1000 kg car on I-5 at 100 km/h, you suddenly noticed a large group of animals. You slam on a break and this provides a stopping force equal to v^3 plus the friction force of 10v. Set up and solve a differential equation for velocity. You need to find a suitable change of variable.
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Zurtex
#2
Jul24-05, 07:59 PM
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What have you attempted?
spoon
#3
Jul24-05, 09:34 PM
P: 12
So far? Nothing...I'm really looking for a place to start =(

Pyrrhus
#4
Jul24-05, 09:44 PM
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Change of variable dif equation problem

Start off with Newton's 2nd Law

[tex] \sum_{i=1}^{n} F_{i} = m \frac{dv}{dt} [/tex]
spoon
#5
Jul24-05, 10:55 PM
P: 12
Alright, so I starting with Newton's equation...I'm thinking:
F = m*V = 1000*V
then I'm assuming F = Stopping force which is: F = V^3 +10V = 1000V
Integrating both sides...
(V^4)/4 + 5V^2 = 500V^2
Then solving for V, I got 13.416
But this totally seems wrong to me...
Pyrrhus
#6
Jul24-05, 11:03 PM
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You cannot integrate it like that

[tex] v^3 + 10v = m \frac{dv}{dt} [/tex]

[tex] \frac{dt}{m} = \frac{dv}{v^3 + 10v} [/tex]
spoon
#7
Jul24-05, 11:25 PM
P: 12
So integrating that I got
t/m = .1 ln|v| - .05 ln|v^2 + 10|
I wasn't really sure how to integrate the right hand side on paper, so I used my calculator
Solving for v, using m = 1000:
e^(t/50) = v/((v^2) + 10)
I used a variable g to = e^(t/50) which led to:
gv^2 +10g = v
I tried using the quadratic after that:
(1 +/- [1-40g]^.5 )/2g = v
now I'm totally confused
qbert
#8
Jul24-05, 11:50 PM
P: 185
Quote Quote by cyclovenom
[tex] v^3 + 10v = m \frac{dv}{dt} [/tex]
If you think about this physically for a sec
you'll see you need a minus sign.
dv/dt really aught to be getting smaller
not bigger.

mv' = -(v^3 + 10v)

this is seperable as above.
[tex] -\frac{dv}{v^3 + 10v} = m dt [/tex]

to integrate it we can breakit into partial fractions
[tex] \frac{1}{x^3 + 10x} = \frac{A}{x} + \frac{Bx + C}{x^2 + 10} [/tex]
where we pick A, B, C to make the equality hold.

Once you've got the easier to integrate sides.
Integrate using the initial conditions.
[tex] \int_{v_0}^v \ldots dv = \int_0^t \ldots dt [/tex]

then invert the formula to solve for v.


[or if we want to look for a change of vars..]
mv' + v^3 + 10v = 0.
so
m(vv') + v^4 + 10v^2 = 0

[tex] \frac{m}{2} \frac{d}{dt}(v^2) + (v^2)^2 + 10 (v^2) = 0 [/tex]

which looks awefully amenable to the substitution u=v^2..
Pyrrhus
#9
Jul25-05, 12:12 AM
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You're right Qbert the aceleration should be negative, but you make a slight mistake, It's [itex] \frac{dt}{m} [/itex].
spoon
#10
Jul25-05, 12:53 AM
P: 12
Quote Quote by qbert
[or if we want to look for a change of vars..]
mv' + v^3 + 10v = 0.
so
m(vv') + v^4 + 10v^2 = 0

[tex] \frac{m}{2} \frac{d}{dt}(v^2) + (v^2)^2 + 10 (v^2) = 0 [/tex]

which looks awefully amenable to the substitution u=v^2..
Following that...
since m = 1000
(500)u d/dt + u^2 + u = 0

Is it a legal operation to factor a "u" out of this equation?
qbert
#11
Jul25-05, 01:33 AM
P: 185
Quote Quote by Cyclovenom
You're right Qbert the aceleration should be negative, but you make a slight mistake, It's [itex] \frac{dt}{m}[/itex]
doh! hoisted by my own petard.
good catch.
qbert
#12
Jul25-05, 01:41 AM
P: 185
Quote Quote by spoon
Following that...
since m = 1000
(500)u d/dt + u^2 + u = 0

Is it a legal operation to factor a "u" out of this equation?
no way! (in this context, d/dt standing alone is meaningless.)

you have to solve the ODE: m du/dt = -(u^2 + 10u).
spoon
#13
Jul25-05, 02:27 AM
P: 12
Okay, so I solving m du/dt = -(u^2 + 10u):
t/500 + C1 = 1/10 ln(u+10) - 1/10 ln(u)
t/50 + C2 = ln((u+10)/u) where C2 = 10*C1
e^(t/50 + C2) = Ce^(t/50)-1 = 10/u
u = 10(Ce^(t/50))^-1
v^2 = 10(Ce^(t/50))^(-1) -----> v = +/-[10(Ce^(t/50))^-1]^(1/2)
But v should just be the negative right?


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