# ODE trouble

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

$$t=\int dt = \int \frac{dx}{dx/dt} = \int \frac{dx}{v(x)}$$

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thanks alot for the hit. It was what i needed to solve the problem

dextercioby
Homework Helper
You have no reason to thank him for.He gave you the erroneous formula for the solving...

$$\frac{dv(t)}{dt}=-u v^{2}(t)$$

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

Daniel.

dextercioby said:
You have no reason to thank him for.He gave you the erroneous formula for the solving...

$$\frac{dv(t)}{dt}=-u v^{2}(t)$$

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!

dextercioby
Homework Helper
And how would you determine $v(x)$,without knowing $v(t)$ and $x(t)$ ???????????

Daniel.

dextercioby
Homework Helper
da_willem said:
$$t=\int dt = \int \frac{dx}{dx/dt} = \int \frac{dx}{v(x)}$$

Please explain the last step of your equation...Why is it $v(x)$ instead of $v(t)$ ...?? Daniel.

dextercioby said:
And how would you determine $v(x)$,without knowing $v(t)$ and $x(t)$ ???????????

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!

dextercioby
Homework Helper

Daniel.

dextercioby said:
Please explain the last step of your equation...Why is it $v(x)$ instead of $v(t)$ ...?? Daniel.

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

$$x(t)=\int v(t) dt$$

And since he has v(x) he can use the definition of velocity v=dx/dt so dt=dx/v and integrate.

dextercioby said:

Daniel.

Why is the formula I gave him not valid?

dextercioby
Homework Helper
Check post number 7.

Daniel.

You can easily see the validity of:

$$t=\int dt = \int \frac{dx}{dx/dt} = \int \frac{dx}{v(x)}$$

By writing it using a Riemann sum. On every interval dx the amount of time it takes him is inversely proportional to the velocity, wich 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...

dextercioby
Homework Helper
The integrands are:
$$\frac{dx}{\frac{dx(t)}{dt}} =\frac{dx}{v(t)}$$

Not v(x).

Daniel.

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

P.P.S.The correct integration would be
$$\int dx=\int v(t)dt$$
which would give u x(t),knowing v(t)...

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

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

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Well.....?

dextercioby
Homework Helper
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 $$t=\int_0 ^{x_1} \frac{dx}{v(x)}$$. Or why not???!!

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

Well maybe that's mathematically not so nice. But given v(x) wouldn't the expression I advised the OP to use work?!

dextercioby
Homework Helper
It's not "mathematically nice",because it's mathematically incorrect.
From what u've considered,by switching fractions,u'd get
$$v(x)=\frac{dx}{dt}$$

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...

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?!]

dextercioby
Homework Helper
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.

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?!

dextercioby
Homework Helper
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...?? :surprised

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:

$$t=\int \frac{dx}{v(x)}$$

????!!!!! And now just try to answer this simple question....

Or is this one too hard to answer? You could PM it to me if you like.