Let the length of link is L = 40cm and AD = Y, OB = X

X^{2} + Y^{2 }= 1 ...(i)

Diff with respect to time, and -ive sign is taken for down word motion of A, when B is moving in +ive direction, we get

2Xdx/dt – 2Ydy/dt = 0XV_{B} – YV_{A} = 0

V_{A} = (X/Y)V_{B} = (Lcosθ/Lsinθ)V_{B} = VB/tanθV_{A}

= 0.8/tan300 = 1.38m/sec ...(iii)

**V**_{A} = 1.38m/sec

Again differentiating equation (2), we get

Xd^{2}x/dt^{2} + (dx/dt)^{2 }– yd^{2}y/dt^{2} – (dy/dt)^{2} = 0

X.a_{B} + (V_{B})^{2} – Y.a_{A }– (V_{A})^{2} = 0

0.4cos 300 X 0.4 – (0.8)^{2} – 0.4sin 300 X a_{A} – (1.38)^{2}= 0

1.38 + 0.64 – 0.2a_{A} – 1.9 = 0

**a**_{A} = 0.6.6m/sec^{2}