Collar A starts from rest at t=0 and moves upward with a constant acceleration
of 3.6 In./s². Collar B moves downward with a constant velocity of 20 in./s.
Determine the time at which the velocity of block Cis zero. (You must provide an answer before
moving on to the next part.)
The time at which the velocity of block Cis zero is



Answer :

Here's how to determine the time at which the velocity of block C is zero, given the motion of collars A and B:

1. Given Information:
- Collar A starts from rest with a constant upward acceleration of [tex]\(3.6 \text{ in/s}^2\)[/tex].
- Collar B moves downward with a constant velocity of [tex]\(20 \text{ in/s}\)[/tex].

2. Understanding Velocities:
- At any time [tex]\( t \)[/tex], the velocity of collar A, denoted as [tex]\( v_A \)[/tex], can be calculated using the formula for uniformly accelerated motion:
[tex]\[ v_A = a_A \cdot t \][/tex]
where [tex]\( a_A \)[/tex] is the acceleration of collar A. Given [tex]\( a_A = 3.6 \text{ in/s}^2 \)[/tex], we have:
[tex]\[ v_A = 3.6 \cdot t \][/tex]
- The velocity of collar B, denoted as [tex]\( v_B \)[/tex], is constant and downward. Given [tex]\( v_B = 20 \text{ in/s}\)[/tex], we consider downward velocity as negative in the coordinate system where upward is positive. Thus:
[tex]\[ v_B = -20 \text{ in/s} \][/tex]

3. Velocity of Block C:
- Block C's velocity is the algebraic difference between the velocities of collars A and B. Hence, the velocity of block C, denoted as [tex]\( v_C \)[/tex], is given by:
[tex]\[ v_C = v_A - v_B \][/tex]
- Substitute the known expressions:
[tex]\[ v_C = (3.6 \cdot t) - (-20) \][/tex]
Simplifying, we get:
[tex]\[ v_C = 3.6 \cdot t + 20 \][/tex]

4. Finding the Time When [tex]\( v_C = 0 \)[/tex]:
- To determine when the velocity of block C is zero, we set [tex]\( v_C = 0 \)[/tex]:
[tex]\[ 0 = 3.6 \cdot t + 20 \][/tex]
- Solving for [tex]\( t \)[/tex]:
[tex]\[ 3.6 \cdot t + 20 = 0 \][/tex]
[tex]\[ 3.6 \cdot t = -20 \][/tex]
[tex]\[ t = \frac{-20}{3.6} \][/tex]

5. Final Calculation:
- Therefore, the time [tex]\( t \)[/tex] at which the velocity of block C is zero is:
[tex]\[ t = -5.555555555555555 \text{ seconds} \][/tex]

So, the time at which the velocity of block C is zero is [tex]\(-5.56\)[/tex] seconds (rounded to two decimal places).