**39**
Pritish Chakraborty
**·**2010-06-29 03:34:33
I asked for the approach yaar...I don't have the answers.

**23**
qwerty
**·**2010-06-29 04:32:30
take B_{3}A_{3} = r , B_{2}A_{2}=2r , B_{1}A_{1}=3r

now take angle B_{1}A_{1}A_{0} = B_{2}A_{2}A_{1}=B_{3}A_{3}A_{2}= a

v_{A1} = d(A_{0}A_{1})/dt = d/dt ( 2 x 3rcosa)

now

\frac{d(A_{1}A_{2})}{dt} = v_{A_{2}}-v_{A_{1}}

so v_{A_{2}}= \frac{d(A_{1}A_{2})}{dt}+ v_{A_{1}}

v_{A_{2}}= \frac{d(2\times 2rcosa)}{dt}+ v_{A_{1}}

\frac{d(A_{2}A_{3})}{dt} = v_{A_{3}}-v_{A_{2}} = v-v_{A_{2}}

v=\frac{d(A_{2}A_{3})}{dt}+v_{A_{2}}

v_{B}=\frac{d(2rsina)}{dt}

now find relation between v ,r , da/dt .

**23**
qwerty
**·**2010-06-29 04:34:36
then u will get the required vel in terms of v .

**39**
Pritish Chakraborty
**·**2010-06-29 09:08:37
Nice! It sort of makes sense now :D

You took A_{0} as your stationary reference point and measured everything from there. To find the bases of those triangles you used trigo in a half triangle then multiplied by 2.

To find the velocity of A_{1} you again used A_{0} as a reference point, by subtracting those specific velocities. Nice one!

**1**
NITW CIVIL
**·**2017-01-07 17:27:08
this is upto A2 consider v at A2