22. A tennis ball travels down a ramp onto a table and at the end of the ramp has a velocity of .75 m/s. The ball is released from rest at the top of the ramp and travels 1.2 meters to the bottom. What is the acceleration as the ball travels down the ramp? Next create a velocity vs. time graph below taking care to label units and numbers properly. Give the equation for the graph in slope-intercept form (y = mx + b).
22. A tennis ball travels down a ramp onto a table and at the end of the ramp has a velocity of .75 m/s. The ball is released from rest at the top of the ramp and travels 1.2 meters to the bottom. What is the acceleration as the ball travels down the ramp? Next create a velocity vs. time graph below taking care to label units and numbers properly. Give the equation for the graph in slope-intercept form (y = mx + b).
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![**Problem Statement:**
A tennis ball travels down a ramp onto a table and at the end of the ramp has a velocity of 0.75 m/s. The ball is released from rest at the top of the ramp and travels 1.2 meters to the bottom. What is the acceleration as the ball travels down the ramp? Next, create a velocity vs. time graph below taking care to label units and numbers properly. Give the equation for the graph in slope-intercept form (y = mx + b).
---
**Graph Description:**
The image includes an empty graph with labeled axes:
- The horizontal axis is meant for time (t) and should be labeled with appropriate units such as seconds (s).
- The vertical axis is meant for velocity (v) and should be labeled with units like meters per second (m/s).
To solve the problem and create the graph, follow these steps:
1. **Calculate Acceleration:**
- Use the kinematic equation: \( v^2 = u^2 + 2as \),
where \( v = 0.75 \, \text{m/s} \) (final velocity), \( u = 0 \, \text{m/s} \) (initial velocity), and \( s = 1.2 \, \text{m} \) (distance).
- Rearrange to solve for \( a \):
\[
a = \frac{v^2 - u^2}{2s} = \frac{(0.75)^2 - 0}{2 \times 1.2} = \frac{0.5625}{2.4} \approx 0.234375 \, \text{m/s}^2
\]
2. **Create the Velocity vs. Time Graph:**
- Plot the graph with time on the x-axis (0 to some time t) and velocity on the y-axis (0 to 0.75 m/s).
- The line will start at the origin (0, 0) and will have a positive slope, representing constant acceleration.
- Use the equation \( v = at \) to calculate points, where \( a \approx 0.234375 \, \text{m/s}^2 \).
3. **Equation of the Graph:**
- In slope-intercept form, the equation of a straight line is \( v = at](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F64953897-0988-4b68-8293-b4cf6ff9ca7f%2Ff4440815-022b-4de8-8bc2-5980de48a69a%2Fg8k9xva_processed.png&w=3840&q=75)
Transcribed Image Text:**Problem Statement:**
A tennis ball travels down a ramp onto a table and at the end of the ramp has a velocity of 0.75 m/s. The ball is released from rest at the top of the ramp and travels 1.2 meters to the bottom. What is the acceleration as the ball travels down the ramp? Next, create a velocity vs. time graph below taking care to label units and numbers properly. Give the equation for the graph in slope-intercept form (y = mx + b).
---
**Graph Description:**
The image includes an empty graph with labeled axes:
- The horizontal axis is meant for time (t) and should be labeled with appropriate units such as seconds (s).
- The vertical axis is meant for velocity (v) and should be labeled with units like meters per second (m/s).
To solve the problem and create the graph, follow these steps:
1. **Calculate Acceleration:**
- Use the kinematic equation: \( v^2 = u^2 + 2as \),
where \( v = 0.75 \, \text{m/s} \) (final velocity), \( u = 0 \, \text{m/s} \) (initial velocity), and \( s = 1.2 \, \text{m} \) (distance).
- Rearrange to solve for \( a \):
\[
a = \frac{v^2 - u^2}{2s} = \frac{(0.75)^2 - 0}{2 \times 1.2} = \frac{0.5625}{2.4} \approx 0.234375 \, \text{m/s}^2
\]
2. **Create the Velocity vs. Time Graph:**
- Plot the graph with time on the x-axis (0 to some time t) and velocity on the y-axis (0 to 0.75 m/s).
- The line will start at the origin (0, 0) and will have a positive slope, representing constant acceleration.
- Use the equation \( v = at \) to calculate points, where \( a \approx 0.234375 \, \text{m/s}^2 \).
3. **Equation of the Graph:**
- In slope-intercept form, the equation of a straight line is \( v = at
Expert Solution
Introduction:
We are given the ball rolling. We are given initial and final velocity. We are also given the displacement. We find the acceleration first. We then give the equation of velocity. We then sketch the graph.
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