Velocity of a particle with some condition

[Govind_Balaji]

Pl Help:Finding velocity using pythagorasTheorem and other velocities.

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Homework Statement

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,

• $\frac{v_1+v_2}{2}$
  
  
  
• \sqrt{v_1v_2}s
  
  
  
• $\frac{v_1v_2}{v_1+v_2}$
  
  
  
• $\sqrt{2}\frac{v_1v_2}{v_1+v_2}$
  
  

Homework Equations

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

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

$v3=\sqrt{2}\frac{v_1v_2}{v_1+v_2}$

 

[adjacent]

Displacement of particle A=2*adjacent side.

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

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

What is A,O and B?
Can you mark those points on the image?

 

[Govind_Balaji]

What is A,O and B?
Can you mark those points on the image?

Sure.
https://scontent-a-iad.xx.fbcdn.net/hphotos-frc3/t1.0-9/10171047_1594387394119260_4471451455186125126_n.jpg

 

[ehild]

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

 

[Govind_Balaji]

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

I never said $$v_1=v_2$$

 

[Govind_Balaji]

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)

 

[ehild]

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

ehild

 

[Govind_Balaji]

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

ehild

Never mind the image. I want the answer.

 

[ehild]

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

 

[Govind_Balaji]

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.

 

[Govind_Balaji]

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

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

 

[ehild]

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

 

[Govind_Balaji]

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.

 

[ehild]

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.

Very good idea!

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

When I continue this, I get errors.

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

\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)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

 

[Govind_Balaji]

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

\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)


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

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.

 

[ehild]

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.

Show what you did.
Expand and simplify the equation

\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)

x₃² cancels.

ehild

 

[Govind_Balaji]

Show what you did.
Expand and simplify the equation

\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)

x₃² cancels.

ehild

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.

 

[Govind_Balaji]

I found it!

<br /> <br /> \\<br /> \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 ) \\<br /> <br /> \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} \\<br /> <br /> \frac{x_1x_2}{2}=\frac{x_3\left(x_1+x_2 \right )}{2\sqrt{2}} \\<br /> <br /> \frac{2\sqrt{2}.x_1x_2}{2}=x_3\left(x_1+x_2 \right ) \\<br /> <br /> \sqrt{2}x_1x_2=x_3\left(x_1+x_2 \right ) \\<br /> <br /> x_3=\sqrt{2}\frac{x_1x_2}{x_1+x_2} \\<br /> <br /> v_3t=\sqrt{2}\frac{x_1x_2t^2}{\left( x_1+x_2\right )t}\\<br /> <br /> v_3=\sqrt{2}\frac{x_1x_2t^2}{\left( x_1+x_2\right )t^2}\\<br /> <br /> v_3=\sqrt{2}\frac{x_1x_2}{ x_1+x_2}<br />

 

[Govind_Balaji]

I am able to type LaTex .See here:

http://www.codecogs.com/latex/eqneditor.php

 

[ehild]

$v_3=\sqrt{2}\frac{x_1x_2}{ x_1+x_2}$

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

ehild

 

[ehild]

I am able to type LaTex .See here:

http://www.codecogs.com/latex/eqneditor.php

Your LaTex is very nice. There is some material to read also here, at PhysicsForums. https://www.physicsforums.com/showpost.php?p=3977517&postcount=3
Also, by clicking on the ∑ sign, you get a menu for Latex Reference.

ehild.

 

[Govind_Balaji]

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

ehild

Typing mistake

Your LaTex is very nice. There is some material to read also here, at PhysicsForums. https://www.physicsforums.com/showpost.php?p=3977517&postcount=3
Also, by clicking on the ∑ sign, you get a menu for Latex Reference.

ehild.

I knew it, but it's difficult because in the link I specified, when you type \frac{, the remaining }{}[\b] automatically appears. Its more easy.

 

[ehild]

I knew it, but it's difficult because in the link I specified, when you type \frac{, the remaining }{}[\b] automatically appears. Its more easy.

The link is really very useful, thank you!

ehild