The
cable at B is pulled downwards at 2 m/s,and is slowing at 3m/s^{2}.
Determine the velocity and acceleration of block A. 1) 0.5 m/s & 3/4 m/s^{2} 2) 1.0 m/s & 4/3 m/s^{2} 3) 1.5 m/s & 2/5 m/s^{2} 4) 4.0 m/s & 4/3 m/s^{2} 

Position
Coordinate Equation The length of first cord is given by 2S_{A} + S_{C} = l_{1} The length of the second cord is S_{E}  S_{C} + S_{B}  S_{C} = l_{2}. Combinig the two equations, 4S_{A} + S_{B} + S_{E} = l_{1} + l_{2} Thus the velocity equation is, 4V_{A }+V_{B} = 0 V_{A} = 0.5 m/s The acceleration equation is, 4a_{A}+a_{B} = 0 a_{A} = 3/4 m/s^{2}. 