00:01
Solving part of this problem so let xa is the position of color a position of color a similarly let xc is position color of block c and let xb is its position color of b.
00:27
Now i can write the expression as d minus xa plus xc minus xa plus 2xc plus xc minus xb is equal to constant c.
00:44
Now differentiating it with respect to time so d -2xa plus so this expression first can be written as d -2xa plus 4xc -xb is equal to constant.
01:03
Now i will just differentiate this equation so you can write the expression as 0 -2dxa by dt plus 4dxc by dt minus dxb by dt is equal to 0.
01:25
Now substitute v for dx by dt or minus 2va plus 4vc minus vb is equal to 0 so 4vc minus 2 va minus vb is equal to 0 let it be equation number 2 and let this be equation number 1 now going forward and differentiate the equation 2 with respect to time so you will get the expression as 4 dvc pi dt t minus 2 dva by dt minus dvb by dt is equal to 0 or 4 ac minus 2 a a minus a b is equal to 0.
02:23
Finally you will get a expression as ac is equal to 1 by 4 multiplication 2 a a plus ab let it be equation number third.
02:36
Now calculating the acceleration of caller b.
02:40
So i will use the formula vb square minus vb not square is equal to 2ab multiplication xb minus xb not.
02:54
So you will get the value of ab as vb vb2 - vb02 by 2xb - xb0 now finally putting the value here.
03:12
So it will be equal to 150 square by 2x700 which is equal to 16 .07 mm per second square...