Answer:
Explanation:
To solve this problem, we'll break down the velocities and accelerations relative to different frames of reference.
1. **Velocity of B relative to A:**
- The velocity of B relative to A is the difference between their individual velocities. Since the velocity of B is given as 2 m/s to the right, and assuming A is stationary or moving with a negligible velocity compared to B, the velocity of B relative to A is 2 m/s to the right.
2. **Acceleration of B relative to A:**
- The acceleration of B relative to A is the difference between their individual accelerations. Given that the constant acceleration of B is 3 m/s² to the right, and assuming A has negligible acceleration compared to B, the acceleration of B relative to A is also 3 m/s² to the right.
3. **Absolute velocity of point C on the cable:**
- Since block B is moving with a velocity relative to A, and point C is attached to B by the cable, the absolute velocity of point C will be the sum of B's velocity relative to A and the velocity of point C relative to B.
- We are given the velocity of B relative to A (2 m/s to the right). Let's denote the velocity of point C relative to B as \( V_{CB} \).
- Therefore, the absolute velocity of point C (\( V_C \)) can be calculated as:
\[ V_C = V_{B/A} + V_{CB} \]
We have the information needed to calculate the absolute velocity of point C. Let's proceed with the calculation.
Let's denote the velocity of point C relative to B as \( V_{CB} \). Since the velocity of B relative to A is 2 m/s to the right, and we need to find the absolute velocity of point C (\( V_C \)), we'll use the formula:
\[ V_C = V_{B/A} + V_{CB} \]
Given:
- Velocity of B relative to A (\( V_{B/A} \)) = 2 m/s to the right
- Acceleration of B relative to A (\( a_{B/A} \)) = 3 m/s² to the right
We need to find \( V_{CB} \), which is the velocity of point C relative to B.
First, let's find the time it takes for B to reach a velocity of 2 m/s given an acceleration of 3 m/s²:
\[ v = u + at \]
\[ 2 = 0 + (3)t \]
\[ t = \frac{2}{3} \text{ seconds} \]
Now, using the equation of motion for B relative to A:
\[ v = u + at \]
\[ V_{B/A} = 0 + (3)(\frac{2}{3}) \]
\[ V_{B/A} = 2 \text{ m/s} \]
This tells us that B reaches a velocity of 2 m/s after \( \frac{2}{3} \) seconds.
Now, let's find the distance traveled by B during this time using the equation of motion:
\[ s = ut + \frac{1}{2}at^2 \]
\[ s = 0 + \frac{1}{2}(3)(\frac{2}{3})^2 \]
\[ s = \frac{1}{2}(3)(\frac{4}{9}) \]
\[ s = \frac{6}{9} \text{ meters} \]
\[ s = \frac{2}{3} \text{ meters} \]
This is the displacement of B relative to A during this time. Now, since point C is attached to B by the cable, point C also moves the same distance relative to B.
\[ s_{CB} = \frac{2}{3} \text{ meters} \]
Now, we can find \( V_{CB} \) using the equation of motion:
\[ v^2 = u^2 + 2as \]
\[ V_{CB}^2 = 0^2 + 2(-9.81)(\frac{2}{3}) \]
\[ V_{CB}^2 = -13.08 \]
\[ V_{CB} = \sqrt{-13.08} \]
\[ V_{CB} = 3.61 \text{ m/s} \]
Now, we can find the absolute velocity of point C (\( V_C \)):
\[ V_C = V_{B/A} + V_{CB} \]
\[ V_C = 2 + 3.61 \]
\[ V_C = 5.61 \text{ m/s} \]
So, the absolute velocity of point C is 5.61 m/s.
