Block C is driven within the vertical channel such that vC

Block C is driven within the vertical channel such that v_C = 0.41 j m/s and a_C = -0.21 j m/s^2. L = 1.5 m and theta = 58 degrees. Find the component of Specify your answer in m/s^2 to 3 decimal places.

Solution

let the distance from A to wall be x = 1.5 cos 58

let the distance from C to ground be y= 1.5 sin 58

X² + Y² = 1.5

2x dx/dt + 2 y dy/dt = 0

2x vA + 2y vC = 0

=> vA = - y vC / x

   vA = vC tan 58 = - 0.41 x tan 58 = - 0.6561 m / s²

therefore velocity of A =0.6561 m/s² to the wall