The motor at C pulls in the cable with an acceleration a c = (3t^2) m/s^2 , wher

Question

The motor at C pulls in the cable with an acceleration ac = (3t^2) m/s^2, where t is in seconds. the motor at D draws in its cable at aD = 5 m/s^2. If both motors start at the same instant from rest when d = 3m, determine (a) the time needed for d = 0, and (b) the velocities of blocks A and B when this occurs.

Answer in the back of the book lists t = 1.07 s, Va = .605 m/s, Vb = 5.33 m/s.

I came up with t = 1.0406 s, Va = 1.13 m/s, Vb = 5.2 m/s. I used the horizontal surface as the distance plane, with the left side of block A set to zero. I set the initial position of block A to zero, and the initial position of block B to 3. This way when block B moves left (negative) and block A moves right (pos), when their positions are equal, d = 0.

a_b (acceleration of block B) = -5 m/s^2.
v_b (velocity of block B) = -5t + C => v_b_initial = 0, so C = 0 => v_b = -5t m/s
d_b (position of B) = - (5/2)t^2 + C => d_b_initial = 3, so C = 3 => d_b = (-(5/2)t^2 + 3) m

a_a (accel of block A) = 3t^2 m/s^2
v_a = t^3 + C, (v_a_initial = 0) => v_a = t^3 m/s
d_a = (1/4)t^4 + C, (d_a_initial = 0) => d_a = (1/4)t^4

d = 0 when d_a = d_b, so:

(1/4)t^4 = -(5/2)t^2 + 3
t^4 = -10t^2 + 12
t^4 + 10t^2 - 12 = 0, t = 1.406 s.

Obviously my answer for the velocities depends on the value for t. So my question is where did I go wrong? Is my assumption incorrect that block A's movement to the right is the exact same as the movement of the rope into the motor at C (defined by accel = 3t^2)?

Obviously block B has constant acceleration (5 m/s^2). So using the constant acceleration equation for velocity ( v = v_0 + a*t), I got v_b (t = 1.07) = 0 - 5(1.07) = - 5.35 m/s, or 5.35 m/s to the left for block B. This only added to my confusion since the book lists the vel of block B as 5.33 m/s.

Explanation / Answer

The acceleration of block a is (3/2)t2, not 3t2, because of the pulley arrangement; the block moves half as fast as the cable.

If you rework the problem that way your quadratic becomes t4 + 20t2 - 24 = 0, and the solution to that is t = 1.0656 seconds, which rounds to your given solution of t = 1.07.

Your confusion about vb is because of rounding -- yes, you are right that vb = -5t, but 5 * 1.0656 = 5.328 m/s.

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