Given: $\omega_{01A}$ =100 rad/s, $O_{1}$A =100 mm=0.1 m

We know that the linear velocity of crank $O_{1}$A,

$v_{01A}$ = $v_{A}$ = $\omega_{01A}$ × $O_{1}$A =100 × 0.1 =10 m/s.

Now let us locate the required instantaneous centres as discussed below:

By measurement , we find that

$I_{13}$A= 910 mm=0.91 m, $I_{13}$B= 820 mm= 0.82m, $I_{15}$B = 130 mm=0.13 m,

$I_{15}$D= 50 mm=0.05 m, $I_{16}$D= 200 mm=0.2 m, $I_{16}$E= 400 mm=0.4m,

Velocity of point E on the belt lever:

Let $v_{E}$ = Velocity of point E on the bell crank lever,

$v_{B}$= Velocity of point B, and

$v_{D}$ = velocity of point D.

We know that

$v_{A}$/ $I_{13}$A = $v_{B}$ / $I_{13}$B ….(Considering Centre $I_{13}$ )

$v_{B}$ = $v_{A}$/ $I_{13}$A $\times$ $I_{13}$B = 10/0.91 $\times$ 0.82 = 9.01 m/s.

$v_{B}$ / $I_{15}$B = $v_{D}$ / $I_{15}$D

$v_{D}$ = $v_{B}$ / $I_{15}$B $\times$ $I_{15}$D = 9.01/0.13 $\times$ 0.05 = 3.46 m/s

Similarly,

$v_{D}$ / $I_{16}$D = $v_{E}$ / $I_{16}$E

$v_{E}$ = $v_{D}$ / $I_{16}$D $\times$ $I_{16}$D = 3.46/0.2 $\times$ 0.4 = 6.92 m/s