ODE trouble

  • Thread starter Suicidal
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  • #1
Suicidal
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I am having trouble with the following problem:

A rocket sled having an initial speed of 150mi/hr is slowed by a channel of water. Assume that, during the braking process, the acceleration a is given by a(v)=-u*v^2, where v is the velocity and u is constant.

a) write the equations of motion in terms of v and x

b) if it requires a distance of 2000 ft to slow the sled to 15 mi/hr, determine the value of u

c) Find the time required to slow the sled to 15 mi/hr


I’m having trouble with part c) because I can’t find a way to get time into the experesion.

In part b I found v(x)=150e^(-u*x) but I don’t know where to go with this to find time

I would really appreciate some help

thanks
 

Answers and Replies

  • #2
da_willem
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[tex]t=\int dt = \int \frac{dx}{dx/dt} = \int \frac{dx}{v(x)}[/tex]
 
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  • #3
Suicidal
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thanks a lot for the hit. It was what i needed to solve the problem
 
  • #4
dextercioby
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You have no reason to thank him for.He gave you the erroneous formula for the solving...

[tex] \frac{dv(t)}{dt}=-u v^{2}(t) [/tex]

Can u find the v(t)...??Then integrate and find x(t).

Daniel.
 
  • #5
da_willem
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dextercioby said:
You have no reason to thank him for.He gave you the erroneous formula for the solving...

[tex] \frac{dv(t)}{dt}=-u v^{2}(t) [/tex]

Can u find the v(t)...??Then integrate and find x(t).

Daniel.

This was not the intention of the question. First he had to rewrite the DE in terms of v and x, and then solve for v(x) not v(t). He already did that and now for part c) has to find the time using his results of a and b. And the formula I gave him was not erroneous!
 
  • #6
dextercioby
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And how would you determine [itex] v(x) [/itex],without knowing [itex] v(t) [/itex] and [itex] x(t) [/itex] ??

Daniel.
 
  • #7
dextercioby
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da_willem said:
[tex]t=\int dt = \int \frac{dx}{dx/dt} = \int \frac{dx}{v(x)}[/tex]

Please explain the last step of your equation...Why is it [itex]v(x) [/itex] instead of [itex] v(t) [/itex] ...?? :rolleyes:

Daniel.
 
  • #8
da_willem
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dextercioby said:
And how would you determine [itex] v(x) [/itex],without knowing [itex] v(t) [/itex] and [itex] x(t) [/itex] ??

Daniel.

a) write the equations of motion in terms of v and x
b) if it requires a distance of 2000 ft to slow the sled to 15 mi/hr, determine the value of u

And he succeeded; he got as a result:

In part b I found v(x)=150e^(-u*x)

It was all in his post!
 
  • #9
dextercioby
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Well,the issue remains...Your advice was wrong...And naturally it couldn't have lead to any VALID result...

Daniel.
 
  • #10
da_willem
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dextercioby said:
Please explain the last step of your equation...Why is it [itex]v(x) [/itex] instead of [itex] v(t) [/itex] ...?? :rolleyes:

Daniel.

Because he calculated v(x) and not v(t) you can't use:

[tex]x(t)=\int v(t) dt[/tex]

And since he has v(x) he can use the definition of velocity v=dx/dt so dt=dx/v and integrate.
 
  • #11
da_willem
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dextercioby said:
Well,the issue remains...Your advice was wrong...And naturally it couldn't have lead to any VALID result...

Daniel.

Why is the formula I gave him not valid?
 
  • #13
da_willem
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You can easily see the validity of:

[tex]t=\int dt = \int \frac{dx}{dx/dt} = \int \frac{dx}{v(x)}[/tex]

By writing it using a Riemann sum. On every interval dx the amount of time it takes him is inversely proportional to the velocity, which he obtained a s a function of x. Summing over all intervals dx gives him the total time.

He already solved the DE to find v(x), so now he can use v(x) instead of v(t).

Try it your way and see if you get the same result...
 
  • #14
dextercioby
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The integrands are:
[tex] \frac{dx}{\frac{dx(t)}{dt}} =\frac{dx}{v(t)} [/tex]

Not v(x).

Daniel.

P.S.There's no need Riemann sum whatsoever.
[tex] v(t)=\frac{dx}{dt}[/tex]=>what I've written above,which of course is an enormity...

P.P.S.The correct integration would be
[tex] \int dx=\int v(t)dt [/tex]
which would give u x(t),knowing v(t)...
 
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  • #15
da_willem
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Ofcourse solving the DE for v(t) is also an option, but not the intention of the question. He already solved the the DE for v(x). Now if somebody asked:

I have a relation which gives the velocity at a point x namely: [tex]v(x)=150e^{-ux}[/tex] where u is known. I want to know the time it takes to go from x=0 to some given point [itex]x_1[/itex]. How do I do this?

Then the answer is [tex]t=\int_0 ^{x_1} \frac{dx}{v(x)}[/tex]. Or why not?!
 
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  • #16
da_willem
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Well...?
 
  • #17
dextercioby
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da_willem said:
Ofcourse solving the DE for v(t) is also an option, but not the intention of the question. He already solved the the DE for v(x). Now if somebody asked:



Then the answer is [tex]t=\int_0 ^{x_1} \frac{dx}{v(x)}[/tex]. Or why not?!

Because it's wrong...Forget about the definite integral.Let's discuss variables in the differentials
U're saying that
[tex] dt=\frac{dx}{v(x)} [/tex]
.Is that right,or am missing something...?? :rolleyes:

Daniel.
 
  • #18
da_willem
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Well maybe that's mathematically not so nice. But given v(x) wouldn't the expression I advised the OP to use work?!
 
  • #19
dextercioby
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It's not "mathematically nice",because it's mathematically incorrect.
From what u've considered,by switching fractions,u'd get
[tex] v(x)=\frac{dx}{dt} [/tex]

which is incorrect...

Daniel.

P.S.So i was right all along,eh...?From my first message i asserted the advice was wrong and hence useless...
 
  • #20
da_willem
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Why isn't this correct? dx/dt is the definition of velocity and he happens to have v as a function of x, not t. And why is the formula I gave him wrong? Is the result he gets for t incorrect? [So is the formula I gave him indeed useless?!]
 
  • #21
dextercioby
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How about if i told you that the 2 sides of the equation depend on different variables??
"x" in the left and "t" in the right...They ought to be constant then,right??

Daniel.
 
  • #22
da_willem
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Well you keep avoiding my question: Is the result he gets for t incorrect?

I'm not a mathematician. I'm educated to be an engineer and lots of em fumbble with differentials like they're just numbers. So maybe my 'derivation' was pretty messy, but what about the result?!
 
  • #23
dextercioby
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I cannot answer the question on HOW THE ORIGINAL POSTER CAME UP WITH THE RESULT...He does...

Daniel.

P.S.Are u implying that an incorrect method would lead to a correct result...??
 
  • #24
da_willem
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He claimed to already have found an expression for v(x) somehow. Then can't you find the time it takes to go from some point to another by evaluating the integral:

[tex]t=\int \frac{dx}{v(x)} [/tex]

?! And now just try to answer this simple question...
 
  • #25
da_willem
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Or is this one too hard to answer? You could PM it to me if you like.
 
  • #26
dextercioby
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All this time i did the calculations (all of them) on both ways...They match...Do you see why??Your formula is still not correct,though,it lacks something...

Daniel.
 
  • #27
da_willem
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Aha, so my equation wasn't so "wrong" after all. It yielded the correct result :rofl: And ofcourse it should. Putting aside the mathematics isn't it totally obvious intuitivly, physically why integrating the inverse of the velocity over a distance yields the time to cover this distance? So your comment on my post instead of 'useless, and incorrect' should have been: 'well ofcourse using that formula in combination with v(x) leads to the correct result but your 'derivation' isn't that neat'.
 
  • #28
dextercioby
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Your formula should have stated:
[tex] \int dt =\int\frac{dx}{v(t(x))} [/tex]

which is different from what u've written...They're basically the same function of "x",but mine explicitely says that the "x" explicit dependence is achived implicitely...

Daniel.
 
  • #29
da_willem
599
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dextercioby said:
Your formula should have stated:
[tex] \int dt =\int\frac{dx}{v(t(x))} [/tex]

which is different from what u've written...They're basically the same function of "x",but mine explicitely says that the "x" explicit dependence is achived implicitely...

Daniel.

You could have said that right away, then I would have agreed, then this thread wouldn't have lasted 28 posts (29 now). So in the futue please don't be so 'mierenneukerig' (I don't know the English word for it) and just correct things instead of saying it's total nonsense...Thanks...
 

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