# Velocity of a particle with some condition

1. May 1, 2014

### Govind_Balaji

Pl Help:Finding velocity using pythagorasTheorem and other velocities.

$\left[$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,

• $\frac{v_1+v_2}{2}$

• $\sqrt{v_1v_2}s$

• $\frac{v_1v_2}{v_1+v_2}$

• $\sqrt{2}\frac{v_1v_2}{v_1+v_2}$

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. May 1, 2014

### adjacent

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

3. May 2, 2014

### Govind_Balaji

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

4. May 2, 2014

### 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

5. May 2, 2014

### Govind_Balaji

I never said $$v_1=v_2$$

6. May 2, 2014

### 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)

7. May 2, 2014

### ehild

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

ehild

#### Attached Files:

• ###### threeparticle.JPG
File size:
6 KB
Views:
73
Last edited: May 2, 2014
8. May 2, 2014

### Govind_Balaji

Never mind the image. I want the answer.

9. May 2, 2014

### ehild

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

ehild

10. May 2, 2014

### 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.

11. May 2, 2014

### 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

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

12. May 2, 2014

### 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

#### Attached Files:

• ###### threepartvel.JPG
File size:
6 KB
Views:
72
13. May 2, 2014

### 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.

Last edited: May 2, 2014
14. May 2, 2014

### ehild

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

15. May 2, 2014

### Govind_Balaji

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.

16. May 2, 2014

### ehild

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

17. May 2, 2014

### Govind_Balaji

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.

18. May 2, 2014

### Govind_Balaji

I found it!

$\\ \frac{x_1x_2}{2}-\frac{x_3^2}{2}=\frac{1}{2}\left ( \frac{x_3x_2}{\sqrt{2}} +\frac{x_3x_1}{\sqrt{2}} - \frac{2x_3^2}{2}\right ) \\ \frac{x_1x_2}{2}-\frac{x_3^2}{2}= \frac{x_3x_2}{2\sqrt{2}} +\frac{x_3x_1}{2\sqrt{2}} - \frac{x_3^2}{2} \\ \frac{x_1x_2}{2}=\frac{x_3\left(x_1+x_2 \right )}{2\sqrt{2}} \\ \frac{2\sqrt{2}.x_1x_2}{2}=x_3\left(x_1+x_2 \right ) \\ \sqrt{2}x_1x_2=x_3\left(x_1+x_2 \right ) \\ x_3=\sqrt{2}\frac{x_1x_2}{x_1+x_2} \\ v_3t=\sqrt{2}\frac{x_1x_2t^2}{\left( x_1+x_2\right )t}\\ v_3=\sqrt{2}\frac{x_1x_2t^2}{\left( x_1+x_2\right )t^2}\\ v_3=\sqrt{2}\frac{x_1x_2}{ x_1+x_2}$

19. May 2, 2014

### Govind_Balaji

20. May 2, 2014

### ehild

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

ehild

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