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spoon

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- Thread starter spoon
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- #1

spoon

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- #2

Zurtex

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What have you attempted?

- #3

spoon

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So far? Nothing...I'm really looking for a place to start =(

- #4

Pyrrhus

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Start off with Newton's 2nd Law

[tex] \sum_{i=1}^{n} F_{i} = m \frac{dv}{dt} [/tex]

[tex] \sum_{i=1}^{n} F_{i} = m \frac{dv}{dt} [/tex]

- #5

spoon

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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...

- #6

Pyrrhus

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

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

- #7

spoon

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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

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

Last edited:

- #8

qbert

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cyclovenom said:[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..

Last edited:

- #9

Pyrrhus

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- #10

spoon

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qbert said:[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?

- #11

qbert

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Cyclovenom said: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.

- #12

qbert

- 185

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spoon said: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).

- #13

spoon

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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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