The cable at B is pulled downwards at 2 m/s,and is slowing at 3m/s2. Determine the velocity and acceleration of block A.

1) 0.5 m/s & -3/4 m/s2
2) 1.0 m/s & -4/3 m/s2
3) 1.5 m/s & 2/5 m/s2
4) 4.0 m/s & 4/3 m/s2
Position Coordinate Equation
The length of first cord is given by
2SA + SC = l1
The length of the second cord is
SE - SC + SB - SC = l2.
Combinig the two equations,
4SA + SB + SE = l1 + l2
Thus the velocity equation is,
4VA +VB = 0
VA = 0.5 m/s
The acceleration equation is,
4aA+aB = 0
aA = -3/4 m/s2.