Velocity relation of a particle

In summary, Daniel finds the time for a particle to reach a certain point by integrating the equation for velocity, given that the mass and initial location of the particle are known. He finds that the time is approximately T= [pi/3 + sqrt(3)/4], which is correct.
  • #1
Wiz
21
0
I have a relation for velocity as v= sqrt [(1-x)/x)]
I need to find the time the particle takes to reach x= 0.25
Initially it is at x=1
I had problems integrating furthur...to find time
Can you help??..
Wiz
 
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  • #2
I did this integral on my calulator and it's gross. Integration by parts perhaps? See what you can do with that..

[tex]\int \sqrt{\frac{1-x}{x}} dx = \int (\sqrt{(1-x)}) * (\sqrt{x})^{-1} dx[/tex]


Jameson
 
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  • #3
From the definition of velocity
[tex] v(t)=:\frac{dx}{dt} [/tex] Separation of variables and corresponding limits of integration give:

[tex] \int_{0}^{\tau} dt=\int_{1}^{0.25} \frac{dx}{\sqrt{\frac{1-x}{x}}} [/tex]

The integral in the first member is really simple,it is the time you need,denoted by me with " \tau"...

For the integral in the RHS,use 2 substitutions

First
[tex] x=t^{2} [/tex]
The second
[tex] t=\sin y [/tex]

Pay attention to:
1.Signs.
2.Transformations of the limits of integration.

Daniel.



EDIT:I'm getting a "tau" smaller than 0.It's weird.Are u sure the problem is stated correctly...?
 
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  • #4
prob

I'm not sure...here's the original prob from which i derived tht velocity relation.

There is a particle initially on x = 1
It is acted upon by a force which varies as F(x)= -0.5 * x^-2
What is the time it takes to reach x= 0.25

Thanks,
Wiz
 
  • #5
forgot

I forgot...the mass of the particle is 10^-2 kg...
 
  • #6
That's another problem,man,totally different.

The advice is the same.Separate variables & integrate with corresponding limits.(EDIT 1:It doesn't work,it's second order).

Daniel.

EDIT 2:See next post for further comments.
 
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  • #7
Nope,i'm afraid this problem (the second u posted) is not completely integrable

It requires solving the ODE

[tex] \frac{d^{2}x(t)}{dt^{2}}=-|C|\frac{1}{x^{2}(t)} [/tex]

,but subject only to the initial condition:

[tex] x(0)=1 [/tex]

Unfortunately,the closure of the Cauchy problem requires 2 independent initial conditions to determine the 2 coefficients (<============ a II-nd order ODE).

Daniel.

P.S.Look it up again.
 
  • #8
The prob i posted is the same,
All i did was...i found out acc. from the force..
Then integrated it and i found velocity...the constant in tht case was c=-1

Btw i hv mentioned tht the particle is initially at x=1
The prob is same as given in the book,
This is an iit-jee question meant for students who hv cleared the 12th grade...and i am in 11th rite now,
Thanks for the help but i think u are getting a bit too advanced for me,,,

Wiz
 
  • #9
Could you please post your work.I'm really curious of what you've done...

Daniel.
 
  • #10
Oh,my god...i am sooooo sorry
the correct eq is F(x)= -k * 0.5 * x^-2
where k= 10^-2 Nm^2
the mass of the body is 10^-2 kg
Now itz 100% correct.

Wht i did was, i got a= -0.5 * x^-2 as k and m canceled out
so i got dv/dt=(-0.5 * x^-2)
hence dv/dx*dx/dt=(-0.5 * x^-2)
hence dv*v=(-0.5 * x^-2)dx
integrating i got c=-1 and simplification led to the vel eq i posted...
Cud u please check it now?
 
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  • #11
It looks correct.(In general,the constant are less worrying).

Daniel.

P.S.Can u uniquely determine v=v(x)...?
 
  • #12
You can be sure. I also got: v= sqrt [(1-x)/x)].
It's time just to integrate. Substitution sqrt [(1-x)/x)]=t leads to rather simple integral.
 
  • #13
Are u sure it's not a

[tex] \int a(x) \ dx=\int v \ dv [/tex]

Plug the values & integrate.The RHS is simple

[tex] \int v \ dv =\frac{1}{2}v^{2}+C [/tex]



Daniel.
 
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  • #14
I hope we shall see it after integration. There must be "tau">0. If we'll have opposite we can just change the sign. Am i right?
 
  • #15
Sorry. i didn't notice edition. I think You are right in both posts
 
  • #16
The acceleration is force divided by mass:

[tex] a(x)=\frac{F(x)}{m}=\frac{-5\cdot 10^{-3} x^{-2}}{10^{-2}}=-0.5\cdot x^{-2} [/tex]

Therefore:

[tex] \int a(x) \ dx=-0.5\int x^{-2} \ dx =0.5 x^{-1} [/tex]

What initial condition does the velocity satisfy...?

Daniel.
 
  • #17
Hey i solved the prob,
The fact we forgot was tht since the body is moving towards the orgin due to -ve value of acc, the correct relation wud be v= - sqrt[(1-x)/x] ----(1)
Now put x = sin^2($)
hence dx/d$ = sin2$
hence dx = sin2$*d$ --------(2)

By (1),
- dx/sqrt[(1-x)/x] = dt
=> -dx /cot($) = dt
By (2)

dt = - sin2$*d$/cot($)
=> dt = -2sin^2($)*d$
Now integrate lhs from 0 to T and rhs from pi/2 to pi/6
So the final ans I got was
T= [pi/3 + sqrt(3)/4 ]
which happens to be correct :smile:

I hope I haven't made any typing mistakes this time..
Thanks for replying...
 

What is the velocity of a particle?

The velocity of a particle is the rate of change of its position with respect to time. It is a vector quantity and is measured in units of distance per time (e.g. meters per second).

How is velocity related to speed?

Velocity and speed are often used interchangeably, but they are actually different quantities. Velocity is a vector quantity that takes into account both the magnitude and direction of motion, while speed is a scalar quantity that only considers the magnitude of motion.

What is the difference between average velocity and instantaneous velocity?

Average velocity is the total displacement of an object divided by the total time taken, while instantaneous velocity is the velocity of an object at a specific moment in time. Average velocity gives an overall picture of an object's motion, while instantaneous velocity captures specific details.

How is velocity related to acceleration?

Acceleration is the rate of change of velocity with respect to time. In other words, it is how much an object's velocity changes over a certain period of time. If an object's velocity is changing, it is considered to be accelerating. If an object's velocity is constant, it is considered to have zero acceleration.

What factors affect the velocity of a particle?

The velocity of a particle can be affected by several factors, including the forces acting upon it, the mass of the particle, and the direction of its motion. Additionally, the medium through which the particle is moving can also impact its velocity, such as air resistance or friction.

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