Page 69- Practice Problem 3.2 Suppose that as the car continue to move, it has velocity components v₂= 3.5 m/s and v, 1.0 m/s at time t,= 3.0s. What are the magnitude and direction of the average acceleration between t₂=2.0 s and t₂= 3.0s? Answer: 4.7m/s, -58⁰ from +x.
Page 69- Practice Problem 3.2 Suppose that as the car continue to move, it has velocity components v₂= 3.5 m/s and v, 1.0 m/s at time t,= 3.0s. What are the magnitude and direction of the average acceleration between t₂=2.0 s and t₂= 3.0s? Answer: 4.7m/s, -58⁰ from +x.
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Only practice problem 3.2
Second photo had background info for problem.
![**Educational Website Transcription:**
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**Solution**
**Set Up**: Figure 3.7 shows our sketch.
**Solve**: For two-dimensional motion, we need the components of the average acceleration, \(a_{x, \text{avg}}\) and \(a_{y, \text{avg}}\), which are the components of the change in velocity, \(\Delta v_x\) and \(\Delta v_y\), divided by the time interval, \(\Delta t\). From the figure, we find the change in the interval \(t_2 - t_1 = (2.5 \, \text{s}) - (2.0 \, \text{s}) = 0.5 \, \text{s}\).
\[
a_{x, \text{avg}} = \frac{\Delta v_x}{\Delta t} = \frac{3.0 \, \text{m/s}}{0.5 \, \text{s}} = 6.0 \, \text{m/s}^2
\]
The time interval for this interval is zero:
\[
\Delta v_y = 0 \, \text{m/s}
\]
For \(a_{y, \text{avg}} = 0 \text{ m/s}^2 \), the vector points only into the x-direction and has a magnitude of 6.0 m/s².
**Reflect:** A vector quantity is defined in terms of its magnitude and direction. Vector quantities, such as acceleration, velocity, and displacement, all require specification in terms of magnitude and direction in a plane.
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**Figure 3.7 Description:**
The diagram is a graph that represents the vectors for velocity in two-dimensional motion on an x-y coordinate plane. It includes:
- A horizontal x-axis labeled in meters (m).
- A vertical y-axis labeled in meters (m).
- Initial position \(t_1 = 2.0 \, \text{s}\) with vectors \(v_{x1} = 4.0 \, \text{m/s}\) and \(v_{y1} = -10.0 \, \text{m/s}\).
- Final position \(t_2 = 2.5 \, \text{s}\) with vectors \(v_{x2} = 3.0 \, \text{m/s}\) and \(v_{](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F28764828-4d1c-44b8-9544-7d79c8b69e6b%2Fc15e2861-044d-4cf5-8861-ddb83a096421%2Fbhfn15_processed.jpeg&w=3840&q=75)
Transcribed Image Text:**Educational Website Transcription:**
---
**Solution**
**Set Up**: Figure 3.7 shows our sketch.
**Solve**: For two-dimensional motion, we need the components of the average acceleration, \(a_{x, \text{avg}}\) and \(a_{y, \text{avg}}\), which are the components of the change in velocity, \(\Delta v_x\) and \(\Delta v_y\), divided by the time interval, \(\Delta t\). From the figure, we find the change in the interval \(t_2 - t_1 = (2.5 \, \text{s}) - (2.0 \, \text{s}) = 0.5 \, \text{s}\).
\[
a_{x, \text{avg}} = \frac{\Delta v_x}{\Delta t} = \frac{3.0 \, \text{m/s}}{0.5 \, \text{s}} = 6.0 \, \text{m/s}^2
\]
The time interval for this interval is zero:
\[
\Delta v_y = 0 \, \text{m/s}
\]
For \(a_{y, \text{avg}} = 0 \text{ m/s}^2 \), the vector points only into the x-direction and has a magnitude of 6.0 m/s².
**Reflect:** A vector quantity is defined in terms of its magnitude and direction. Vector quantities, such as acceleration, velocity, and displacement, all require specification in terms of magnitude and direction in a plane.
---
**Figure 3.7 Description:**
The diagram is a graph that represents the vectors for velocity in two-dimensional motion on an x-y coordinate plane. It includes:
- A horizontal x-axis labeled in meters (m).
- A vertical y-axis labeled in meters (m).
- Initial position \(t_1 = 2.0 \, \text{s}\) with vectors \(v_{x1} = 4.0 \, \text{m/s}\) and \(v_{y1} = -10.0 \, \text{m/s}\).
- Final position \(t_2 = 2.5 \, \text{s}\) with vectors \(v_{x2} = 3.0 \, \text{m/s}\) and \(v_{
Transcribed Image Text:**Chapter 3: Practice Problems**
**Name:** ____________________________
**G#:** _______________________________
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**Page 66 - Practice Problem 3.1**
Suppose you reverse the car’s motion, so that it retraces its path in the opposite direction in the same time. Find the components of the average velocity of the car and the magnitude and direction of the average velocity.
*Answer: -6.0 m/s, -8.0 m/s, -10.0 m/s, 233°*
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**Page 69 - Practice Problem 3.2**
Suppose that as the car continues to move, it has velocity components \( v_x = 3.5 \) m/s and \( v_y = 1.0 \) m/s at time \( t_3 = 3.0 \) s. What are the magnitude and direction of the average acceleration between \( t_2 = 2.0 \) s and \( t_3 = 3.0 \) s?
*Answer: 4.7 m/s\(^2\), -58° from +x.*
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**Page 73 - Practice Problem 3.3**
If air resistance is ignored, what initial speed is required for a range of 20 m?
*Answer: 36.1 m/s.*
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**Page 74 - Practice Problem 3.4**
If the ball could continue to travel below its original level (through an appropriately shaped hole in the ground), then negative values of \( y \) corresponding to times greater than 6.04 s would be possible. Compare the ball’s position and velocity 8.00 s after the start of its flight.
*Answers: \( x = 178 \) m, \( y = -76.8 \) m, \( v_x = 22.2 \) m/s, \( v_y = -48.8 \) m/s.*
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**Page 75 - Practice Problem 3.5**
Show the range is the same when the launch angle is 30° as when it is 60° and that the range for any launch angle \( \theta \) is the same as for the complementary angle (90° - \( \theta \)).
*Answer: \( \sin \theta = \sin (180° - \theta), \) so \( \sin 2\theta = \sin 2
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