Yet another FODE question

Hey.
I've been doing more mechanics recently - further kinematics in M3. I've come across another question i'm confused by.

A particle moves on the positive x-axis. When its displacement from O is x meters, is acceleration is of magnitude x^-3 m/s^2 and directed towards O. Given that initially, when t=0, the particle is at rest with x=1, find the time the particle takes to reach x=1/4.

First i integrate with respect to x to find:

V^2 = x^-2 + C where C = -1
v^2 = x^-2 -1

V = x^-1 - 1

Now i integrate again to find velocity as (dx/dt):

dx/dt = x^-1 - 1

Which is a FODE, however i cant solve it.

Thank you,
Bob

LeonhardEuler
Gold Member
mr bob said:
Hey.
I've been doing more mechanics recently - further kinematics in M3. I've come across another question i'm confused by.

A particle moves on the positive x-axis. When its displacement from O is x meters, is acceleration is of magnitude x^-3 m/s^2 and directed towards O. Given that initially, when t=0, the particle is at rest with x=1, find the time the particle takes to reach x=1/4.

First i integrate with respect to x to find:

V^2 = x^-2 + C where C = -1
v^2 = x^-2 -1

V = x^-1 - 1

Now i integrate again to find velocity as (dx/dt):

dx/dt = x^-1 - 1

Which is a FODE, however i cant solve it.

Thank you,
Bob
That isn't quite right. The problem says the accleration is given by $x^{-3}$ towards the origin, so
$$\frac{d^2x}{dt^2}=-x^{-3}$$
Multiplying by x3dt2 gives:
$$x^3d^2x =-dt^2$$
Now integrate twice.

HallsofIvy
Homework Helper
LeonhardEuler said:
That isn't quite right. The problem says the accleration is given by $x^{-3}$ towards the origin, so
$$\frac{d^2x}{dt^2}=-x^{-3}$$
Multiplying by x3dt2 gives:
$$x^3d^2x =-dt^2$$
Now integrate twice.

No, you can't treat a second derivative like that, "differentials", dx and dy separately, are only defined for the first derivative: we could separate $\frac{dv}{dt}= x^{-3}$ into $cv= x^{-3}dt$ but then you can't integrate x-3 with respect to t.

You can do this by "quadrature" (looks to me like this is what mr bob did)- since t does not appear explicitly, let $v= \frac{dx}{dt}$. Then $\frac{d^2x}{dt^2}= \frac{dv}{dt}$ and, by the chain rule, $\frac{dv}{dt}= \frac{dv}{dx}\frac{dx}{dt}= v\frac{dv}{dx}$

That makes the equation $v\frac{dv}{dx}= x^{-3}$ which we can separate: $v dv= x^{-3}dx$ and, integrating, $\frac{1}{2}v^2= -\frac{1}{2}x^{-2}+ C$ or $v^2= -x^{-2}+ C$ almost what mr bob said (he is missing the "-" as x increases the particle slows). Taking x= 1, v= 0 when t= 0, we get 0= -1+ C so C= 1 and $v^2= -x^{-2}+ C$ again almost as mr bob said.

However, (are you listening mr bob?) taking the square root does NOT give v= 1- x-1!!! $\sqrt{a^2+ b^2} \ne a+ b$!!!

Instead $v= \frac{dx}{dt}= \sqrt{1- x^{-1}}= \sqrt{\frac{x^2- 1}{x^2}}= \frac{\sqrt{x^2-1}}{x}$. That separates as $\frac{xdx}{x^2- 1}= dt$ which should be easy to integrate.

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HallsofIvy
Homework Helper
LeonhardEuler said:
That isn't quite right. The problem says the accleration is given by $x^{-3}$ towards the origin, so
$$\frac{d^2x}{dt^2}=-x^{-3}$$
Multiplying by x3dt2 gives:
$$x^3d^2x =-dt^2$$
Now integrate twice.

No, you can't treat a second derivative like that: we could separate $\frac{dv}{dt}= x^{-3}$ into $cv= x^{-3}dt$ but then you can't integrat x-3 with respect to t.

You can do this by "quadrature"- since t does not appear explicitly, let $v= \frac{dx}{dt}. Then [itex]\frac{d^2x}{dt^2}= \frac{dv}{dt}$ and, by the chain rule, $\frac{dv}{dt}= \frac{dv}{dx}\frac{dx}{dt}= v\frac{dv}{dx}$

That makes the equation $v\frac{dv}{dx}= x^{-3}$ which we can separate: $v dv= x^{-3}dx$ and, integrating, $\frac{1}{2}v^2= -\frac{1}{2}x^{-2}+ C$ or $v^2= -x^{-2}+ C$ almost what mr bob said (he is missing the "-" as x increases the particle slows). Taking x= 1, v= 0 when t= 0, we get 0= -1+ C so C= 1 and $v^2= 1 -x^{-2}$ again almost as mr bob said.

However, (are you listening mr bob?) taking the square root does NOT give v= 1- x-1!!! $\sqrt{a^2+ b^2} \ne a+ b$!!!

Instead $v= \frac{dx}{dt}= \sqrt{1- x^{-1}}= \sqrt{\frac{x^2- 1}{x^2}}= \frac{\sqrt{x^2-1}}{x}$. That separates as $\frac{xdx}{\sqrt{x^2- 1}}= dt$ which should be easy to integrate.

Thank you very much. I really appreciate the help. Some of these mechanics questions are a bit hard. Need to improve my integration skills abit.

I've integrated to $t = \sqrt(x^2 - 1)$ but cant get the answer correct. Find t when x = 1/4, which gives the root of a negative number, not real. But the book's answer is $\sqrt(15)$ which i notice is $\sqrt(x^-2 - 1)$. How is this possible?

Thank you,
Bob

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HallsofIvy
Homework Helper
The acceleration is of magnitude x-3 and directed toward O!!

I missed that myself: the differential equation is not
$$\frac{dv}{dt}= x^{-3}$$
it is
$$\frac{dv}{dt}= -x^{-3}$$
!!!

That's why you had what I (mistakenly) thought was a sign error!

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When i integrate i get $t = 1 - (1 - x^2)$.
I'm getting confused by this now, i think my brain's just died, ah well. Sorry for keeping asking questions, i can see how annoying it would be if the answer was so obvious yet i dont grasp it.

Thanks,
Bob