Determine the velocity of the piston A at the given instant.

Angular Velocity of the rod CB,

$ω_{CB} = \dfrac {2 πN}{60} = \dfrac {2 ×3.14 ×30}{60} \\ ω_{CB}=3.14 rad/s \\ V_B = CB × ω_{CB}\\ =100 ×3.14 \\ =314 mm/s $

I is obtained by taking perpendicular from $V_A$ and $V_B$

In $∆ ABC \\ \dfrac {AB}{\sin120} = \dfrac {BC}{\sin BAC} = \dfrac {AC}{\sin ABC} \\ \dfrac {400}{\sin⁡120} = \dfrac {100}{\sin ⁡BAC } \\ ∠BAC=12.50^0 \\ ∠ABC=180-120- 12.50^0=47.5^0 \\ \dfrac {400}{\sin ⁡120} = \dfrac {AC}{\sin⁡47.5} \\ AC=340.53 mm \\ In\space\space ∆ ACI (∆ ACI \space is\space 30^0-60^0- 90^0\space triangle ) \\ \dfrac {AC}{\sin30} = \dfrac {AI}{\sin60} = \dfrac {CI}{\sin90 } \\ \dfrac {340.53}{\sin⁡30} = \dfrac {AI}{\sin⁡60} = \dfrac {CI}{\sin⁡90 } \\ AI=589.8 \space\space\space\space\space\space CI=681.06 \\ EI=100+681.06 \\ =781.06 $

Angular velocity of the rod AB,

$ω_{AB} = \dfrac {V_B}{BI} = \dfrac {314}{781.06} =0.402 rad/sec $

Instantaneous velocity of A,

$V_A=r.ω_{AB} = IA.ω_{AB} \\ =589.8 ×0.402 \\ V_A=237.2 mm/s$