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Sagot :
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).
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).
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