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Velocity of a particle with some condition

  1. May 1, 2014 #1
    Pl Help:Finding velocity using pythagorasTheorem and other velocities.

    [itex]\left[[/itex]1. The problem statement, all variables and given/known data
    Three particles start moving simultaneously from a point on a horizontal smooth plane.
    First particle moves with speed ##v_1## towards east, second particle moves towards north with speed ##v_2## and third one moves towards north east. The velocity of the third particle, so that three particles always lie on a straight line is,

    • ##\frac{v_1+v_2}{2}##


    • [itex]\sqrt{v_1v_2}s[/itex]


    • ##\frac{v_1v_2}{v_1+v_2}##


    • ##\sqrt{2}\frac{v_1v_2}{v_1+v_2}##


    2. Relevant equations

    https://fbcdn-sphotos-b-a.akamaihd.net/hphotos-ak-frc3/t1.0-9/10271500_1594365894121410_7484216690186519659_n.jpg

    3. The attempt at a solution

    Let ##v_1=\frac{x_1}{t}##, ##v_2=\frac{x_2}{t}##, ##x_3=\frac{x_3}{t}##, where ##x_1, x_2, x_3## are displacement of three particles in ##t## seconds respectively.

    ##x_1=v_1t##

    ##x_2=v_2t##

    ##x_3=v_3t##

    Displacement of third particle is the hypotenuse of the orange rectangle in the figure.

    Displacement of particle A=2*adjacent side.

    OB=Adjacent Side=##\frac{x_1}{2}##

    Similarly, AB=Opposite side=##\frac{x_2}{2}##

    ##OA^2=x_3^2=\left[\frac{x_1}{2}\right]^2+\left[{\frac{x_2}{2}}\right]^2##

    ##x_3=\sqrt{\frac{x_1^2}{4}+\frac{x_2^2}{4}}##


    ##x_3=\frac{\sqrt{x_1^2+x_2^2}}{\sqrt{4}}##


    ##x_3=\frac{\sqrt{v_1^2t^2+v_2^2t^2}}{2}##


    ##x_3=\frac{\sqrt{(v_1^2+v_2^2)t^2}}{2}##


    ##x_3=\frac{t\sqrt{(v_1^2+v_2^2)}}{2}##


    ##v_3t=\frac{t\sqrt{(v_1^2+v_2^2)}}{2}##


    ##v_3=\frac{\sqrt{(v_1^2+v_2^2)}}{2}##


    I checked my answer by substituting values for ##v_1## and ##v_2##. It's right.

    I assume option D must be correct because it is the only answer which has the term 2.

    I checked and D is the right answer but I should take

    ##v_3=\frac{\sqrt{(v_1^2+v_2^2)}}{2}##

    to

    ##v3=\sqrt{2}\frac{v_1v_2}{v_1+v_2}##
     
    Last edited: May 2, 2014
  2. jcsd
  3. May 1, 2014 #2

    adjacent

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    What is A,O and B?
    Can you mark those points on the image?
     
  4. May 2, 2014 #3
    Sure.
    https://scontent-a-iad.xx.fbcdn.net/hphotos-frc3/t1.0-9/10171047_1594387394119260_4471451455186125126_n.jpg
     
  5. May 2, 2014 #4

    ehild

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    v1 is not equal to v2, but the third particle moves to north-east, so the orange triangle is isosceles.
    The picture is very nice, but you should modify a bit.



    ehild
     
  6. May 2, 2014 #5
    I never said $$v_1=v_2$$
     
  7. May 2, 2014 #6
    Pls help me take

    I checked and D is the right answer but I should take

    v3=√(v1^2+v2^2)/2

    to

    v3=√2(v1v2)/(v1+v2)
     
  8. May 2, 2014 #7

    ehild

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    Draw the track of the third particle in the direction of north-east.

    ehild
     

    Attached Files:

    Last edited: May 2, 2014
  9. May 2, 2014 #8
    Never mind the image. I want the answer.
     
  10. May 2, 2014 #9

    ehild

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    The rules of the forum do not allow me to give the answer.


    ehild
     
  11. May 2, 2014 #10
    No I don't want the exact answer. Just give me a hint. Wait I will work out for a while. If I can't do, then give me an hint. Now don't give me hint.
     
  12. May 2, 2014 #11
    https://fbcdn-sphotos-b-a.akamaihd.net/hphotos-ak-frc3/t1.0-9/10309196_1578688112357075_4699213260430609082_n.jpg

    With the help of your image and some modifcations,

    I got close but not right answer.

    I got

    $$v_3= - \sqrt{2}\frac{v_1v_2}{v_3-v_1+v_2}$$

    instead of

    $$v_3= \sqrt{2}\frac{v_1v_2}{v_1+v_2}$$
     
  13. May 2, 2014 #12

    ehild

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    You need to have v3 on one side only. I can not find where you went wrong if you do not show your work.
    See attachment. The two yellow triangles are similar. The green square have sides equal to both the x and y components of v3. What are the sides of the yellow triangles? What equation you get from the similarity?


    ehild
     

    Attached Files:

  14. May 2, 2014 #13
    Side of the square with $x_3$ as diagonal=a

    $$a^2+a^2=x_3^2$$
    $$2a^2=x_3^2$$
    $$\sqrt{2a^2}=x_3$$
    $$\sqrt{2}a=x_3$$
    $$a=\frac{x_3}{\sqrt{2}}$$


    $$Area of the square=$a^2$$
    $$=\left(\frac{x_3}{\sqrt{2}}\right)^2$$
    $$=\frac{x_3^2}{2}$$



    Area of the triangle fromed by $x_1$ and $x_2$ - Area of the square=

    Area of the triangle formed by 2 small triangles.

    $$=\frac{x_1x_2}{2}-\frac{x_3}{\sqrt{2}}=$$
    $$\frac{1}{2}\left(\frac{x_3}{\sqrt{2}}\right)\left(x_2+x_1-\frac{x_3}{\sqrt{2}}\right)$$.

    When I continue this, I get errors.
     
    Last edited: May 2, 2014
  15. May 2, 2014 #14

    ehild

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    Very good idea!

    Well, there are errors in the previous lines.

    You forgot a square: the area of the little square is ##\frac{x_3^2}{2}##

    The sides of the small triangles are x1-x3/√2 and x2-x3/√2, so you have to subtract x3/√2 twice.

    The correct equation is

    [tex]\frac{x_1x_2}{2}-\frac{x_3^2}{2}=\frac{1}{2}\left(\frac{x_3}{\sqrt{2}}\right)\left(x_2+x_1-2\frac{x_3}{\sqrt{2}}\right)[/tex]


    It would have been simpler to equate the area of the big triangle and the sum of the other two, with bases x1 andx2 and hight x3/√2.

    ehild
     
  16. May 2, 2014 #15
    I get upto this. After this, I don't know where to go

    ##v_3^2=\frac{-v_1v_3-v_2v_3-\sqrt{2}v_1v_2}{2}##

    Pls help.
     
  17. May 2, 2014 #16

    ehild

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    Show what you did.
    Expand and simplify the equation

    [tex] \frac{x_1x_2}{2}-\frac{x_3^2}{2}=\frac{1}{2}\left(\frac{x_3}{\sqrt{2}}\right)\left(x_2+x _1-2\frac{x_3}{\sqrt{2}}\right)[/tex]

    x32 cancels.

    ehild
     
  18. May 2, 2014 #17
    OH! I missed that x_3^2/2 cancles.

    I will work out and see. I feel tired of typing long and complex latex. that's why I am not able to type all my work.
     
  19. May 2, 2014 #18
    I found it!

    [itex]

    \\
    \frac{x_1x_2}{2}-\frac{x_3^2}{2}=\frac{1}{2}\left ( \frac{x_3x_2}{\sqrt{2}} +\frac{x_3x_1}{\sqrt{2}} - \frac{2x_3^2}{2}\right ) \\

    \frac{x_1x_2}{2}-\frac{x_3^2}{2}= \frac{x_3x_2}{2\sqrt{2}} +\frac{x_3x_1}{2\sqrt{2}} - \frac{x_3^2}{2} \\

    \frac{x_1x_2}{2}=\frac{x_3\left(x_1+x_2 \right )}{2\sqrt{2}} \\

    \frac{2\sqrt{2}.x_1x_2}{2}=x_3\left(x_1+x_2 \right ) \\

    \sqrt{2}x_1x_2=x_3\left(x_1+x_2 \right ) \\

    x_3=\sqrt{2}\frac{x_1x_2}{x_1+x_2} \\

    v_3t=\sqrt{2}\frac{x_1x_2t^2}{\left( x_1+x_2\right )t}\\

    v_3=\sqrt{2}\frac{x_1x_2t^2}{\left( x_1+x_2\right )t^2}\\

    v_3=\sqrt{2}\frac{x_1x_2}{ x_1+x_2}
    [/itex]
     
  20. May 2, 2014 #19
  21. May 2, 2014 #20

    ehild

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    Yes, but write v1,v2 instead of x1,x2.

    ehild
     
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