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Tangent help please

  • Thread starter ritwik06
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  • #1
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Homework Statement





Homework Equations





The Attempt at a Solution

 

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  • #2
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Sorry guys. I posted an empty thread by mistake due to power failure. Here is the question.

Homework Statement



Point A moves uniformly with velocity v so that the vector v is continually "aimed" at point B which in its turn moves rectilinearly and uniformly ith velocity u <v. At the initial moment of time v is perpendicular to u and the points are separatd by distance k. How soon will the points meet?



The Attempt at a Solution

Refer to diagram:

All that I could make out from the question was that the path followed by the Point A will be a parabola. And the angles between vectors will continually change, it will decrase to be specific.

A is assumed to be at the origin of the cartesian plane.

At any instance of time (t):
Velocity of A relative to B
V(X component)= v cos [tex]\vartheta[/tex]
V(Y component)= v sin [tex]\vartheta[/tex]-u
[tex]\vartheta[/tex] is the angle which the velocity vector v makes with x axis. [tex]\vartheta[/tex] changes with time.

Acceleration:
A(x)=-v sin [tex]\vartheta[/tex]*(d[tex]\vartheta[/tex]/dt)
A(y)=v cos [tex]\vartheta[/tex]*(d[tex]\vartheta[/tex]/dt)
magnitude of acceleration(as obtained from the above components)= v *(d[tex]\vartheta[/tex]/dt)
=v [tex]\omega[/tex]
[tex]\omega[/tex]is the angular velocity.


Well, I havent studied kinematics in very much details, not even circular motion. But I am able to solve simple questions on circular motion. I know simple usage of basic calculus. Please guide me to solve this, as its been a long time since i hav been trying this question.
 

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  • #3
tiny-tim
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Sorry guys. I posted an empty thread by mistake due to power failure
:biggrin: ah … and I thought your dog ate it! :biggrin:
Well, I havent studied kinematics in very much details, not even circular motion. But I am able to solve simple questions on circular motion. I know simple usage of basic calculus. Please guide me to solve this, as its been a long time since i hav been trying this question.
You don't need "kinematics", or even acceleration, just a clear head (and enough light to see the problem!).

You simply need to know what θ is (or tanθ) …

and you can work that out because you know exactly where B is at time t, and you just subtract (x,y) from that. :smile:
 
  • #4
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:biggrin: ah … and I thought your dog ate it! :biggrin:


You don't need "kinematics", or even acceleration, just a clear head (and enough light to see the problem!).

You simply need to know what θ is (or tanθ) …

and you can work that out because you know exactly where B is at time t, and you just subtract (x,y) from that. :smile:
Yes, I know where point B is after time "t". It will travel "ut" in positive direction of x axis.
But how can tan θ be determined from that???:confused: Actually please tell me what do ya mean by (x,y). What coordinates are these? I think tan θ at any instant t will vary:

The perpendicular would be distance traversed by B (ie, ut) in that time - the vertical distance covered by A in that duration.
Similarly base would be the remaining horizontal separation.
Now how can these things be determined?
θ is the angle made by vector v with x axis. Is it not?
Please help me. I am new to these things. And i really dont have enough patience to go one with just on problem. its simply infuriating... :) thx for the help
 
Last edited:
  • #5
tiny-tim
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Hi ritwik06! :smile:

(x,y) is the position of A … x along the x-axis, and y up the y-axis.

As you say, B is at position (k,ut).

So the vector AB is (k - x, ut - y), and so tanθ = (ut - y)/(k - x). :smile:
 
  • #6
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Hi ritwik06! :smile:

(x,y) is the position of A … x along the x-axis, and y up the y-axis.

As you say, B is at position (k,ut).

So the vector AB is (k - x, ut - y), and so tanθ = (ut - y)/(k - x). :smile:
Hi tim~
But u have added two more unknowns to my equation :smile:

Most probably I have to diffrentiate the equation obtained. right? If yes, with respct to what ?? There are 4 variables in the function x,y,θ,t...

how shall i proceed? Please help me dear. I dont really have any idea about this. I had tried my best and shown u the results.

And on more thing:
when the points meet ut=y
and k=x
therefore tan θ is undefined

Please do help me clearly tim. Do u want me to remain stuck with this?
 
  • #7
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Somebody help me please. Tim r u there?
 
  • #8
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so nobody could solve it right now, eh?
 
  • #9
tiny-tim
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Hi ritwik06! :smile:

You already have all the information you need to solve this.

You know that the magnitude of v is constant.

You know the direction of v as a function of x y and t.

So write out the differential equations, and solve them! :smile:
 
  • #10
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Hi tim;
This is Ritwik.
Well the thing is that I have been trying to solve that question for a long time now.

As u say, "v" is constant. But "v" is not involved in our equation for tan [tex]\theta[/tex]. I tried to differentiate the equation in tan [tex]\theta[/tex] with respect to time t:
I got this:

u=(sec^2 [tex]\theta[/tex]) *[tex]\omega[/tex]*(k-x)

There is still the angular velocity and x coordinate.

And please I would like to tell u one thing. I am in standard 11 only. I have not even learned calculus fully. Just some basic formulae.

Please will u help me with this through?
Thanks for all the effort!
Ritwik
 
  • #11
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Kinematics

Point A moves uniformly with velocity v so that the vector v is continually "aimed" at point B which in its turn moves rectilinearly and uniformly with velocity u <v. At the initial moment of time v is perpendicular to u and the points are separatd by distance k. How soon will the points meet?


All that I could make out from the question was that the path followed by the Point A will be a parabola. And the angles between vectors will continually change, it will decrase to be specific.

A is assumed to be at the origin of the cartesian plane.

At any instance of time (t):
Velocity of A relative to B
V(X component)= v cos [tex]\vartheta[/tex]
V(Y component)= v sin [tex]\vartheta[/tex]-u
[tex]\vartheta[/tex] is the angle which the velocity vector v makes with x axis. [tex]\vartheta[/tex] changes with time.

Acceleration:
A(x)=-v sin [tex]\vartheta[/tex]*(d[tex]\vartheta[/tex]/dt)
A(y)=v cos [tex]\vartheta[/tex]*(d[tex]\vartheta[/tex]/dt)
magnitude of acceleration(as obtained from the above components)= v *(d[tex]\vartheta[/tex]/dt)
=v [tex]\omega[/tex]
[tex]\omega[/tex]is the angular velocity.


Well, I havent studied kinematics in very much details, not even circular motion. But I am able to solve simple questions on circular motion. I know simple usage of basic calculus. Please guide me to solve this, as its been a long time since i hav been trying this question.
 
Last edited:
  • #12
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So nobody could answer this here as well????
 
  • #13
malawi_glenn
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you have not done an attempt to solution.
 
  • #15
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you have not done an attempt to solution.
Well, I have shown all that I could. I have been trying to solve this for a long time. I have tried my best. I had even put it in introductory physics. The people who tried to help me did not have their concepts clear. I can even show you a PM from one of the members getting himself stuck at last and rgretting. Its now upto u whether u ish to help or not, of course if u know it urself.
 
  • #16
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  • #17
ZapperZ
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Both threads on the identical problem have been merged. Please do not start a new thread on this. If it drops too far down the list, simply bump it back up.

However, I think you were given ample information to solve this assuming that you have the necessary mathematics. If not, I'd say that the problem isn't with this question, but the lack of skill to solve it.

Zz.
 
  • #18
malawi_glenn
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"And the rest is all upon u guys."

Never blaim us for your inability to solve a problem!!!
 
  • #19
malawi_glenn
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so nobody could solve it right now, eh?
We NEVER solve problems here, we HELP people solving problems by giving hints and pushing in right direction.
 
  • #20
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MODS; its no use. close this thread. I have got my problem solved.
 

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