Solving Collar A, B, and C's Motion

[385sk117]

Homework Statement

Collar A starts from rest at t=0 and moves upward with a constant accleration of 9cms^-2. Knowing that collar B moves downward with a constant velocity of 45.7cms^-1, determine the time at which the velocity of block c is zero and the corresponding position of block c

Homework Equations

Xa + Xb + Xc = constant

dXa/dt + dXb/dt + dXc/dt = 0

Va + Vb + Vc = 0

dVa/dt + dVb/dt + dVc/dt = 0

Aa + Ab + Ac = 0

Aa = 9cms^-2 constant
Va = 9t cms^-1
Xa = 4.5t^2 + Xao cm

Ab = 0
Vb = -45.7cms^-1
Xb = -45.7t + Xbo cm

The Attempt at a Solution

At first I thought that 9t - 45.7 = 0 is the equation to find the time t when the velocity of block C is 0, but the picture confused me as it shows like when B goes down , block C will go up and when A goes up block C will go up as well, so the block will keep going up, won't it be like that? and even if i found out the time t, the position of block C can not be found as the initial positions of A, B, and C are unclear.
Help me please TT

 

[darkdream]

my 5 cents advice: look at the diagram in terms of rope distance.

the traveling path of C will be something like:
the ropes on the right side will lossen at the rate of twice of the velocity of collar B.

This makes C falls.

the rope on the left connecting C to A will pull up C at a rate of 1/2 of wadever distance gained since A only pulls on one side of the rope.

This makes C raise.

Hope it helps

 

[385sk117]

Thanks a lot!