A baseball is pitched, reaching the batter with a speed of 37 m/s, and is hit such that it returns on the same line at 48 m/s. If the ball was in contact with the bat for 0.22 seconds, what was the magnitude of the average acceleration experienced by the ball during the time of contact?
A baseball is pitched, reaching the batter with a speed of 37 m/s, and is hit such that it returns on the same line at 48 m/s. If the ball was in contact with the bat for 0.22 seconds, what was the magnitude of the average acceleration experienced by the ball during the time of contact?
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Chapter1: Units, Trigonometry. And Vectors
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![## Physics: Calculating Average Acceleration
### Problem Statement
A baseball is pitched, reaching the batter with a speed of 37 m/s, and is hit such that it returns on the same line at 48 m/s. If the ball was in contact with the bat for 0.22 seconds, what was the magnitude of the average acceleration experienced by the ball during the time of contact?
### Multiple Choice Options
- ○ 220 m/s²
- ○ 390 m/s²
- ○ 41 m/s²
- ○ 170 m/s²
### Solution Steps
To find the magnitude of the average acceleration, use the formula:
\[ a = \frac{\Delta v}{\Delta t} \]
where \( \Delta v \) is the change in velocity and \( \Delta t \) is the time duration.
Let's calculate it step by step:
1. **Initial velocity (u)**: 37 m/s (as the ball was pitched towards the batter)
2. **Final velocity (v)**: 48 m/s (as the ball returns on the same line)
3. **Time of contact (t)**: 0.22 seconds
Change in velocity (\( \Delta v \)) is given by:
\[ \Delta v = v - (-u) = 48 - (-37) = 48 + 37 = 85 \, \text{m/s} \]
Average acceleration (\( a \)) is:
\[ a = \frac{85 \, \text{m/s}}{0.22 \, \text{s}} \approx 386.36 \, \text{m/s}^2 \]
Therefore, the closest correct option is \( 390 \, \text{m/s}^2 \).](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F706abf0d-1664-42d8-b2e3-1006282c8cfe%2F304ca915-2f8f-4adf-809e-ae2f14a6d12b%2Fe5vqf9s.jpeg&w=3840&q=75)
Transcribed Image Text:## Physics: Calculating Average Acceleration
### Problem Statement
A baseball is pitched, reaching the batter with a speed of 37 m/s, and is hit such that it returns on the same line at 48 m/s. If the ball was in contact with the bat for 0.22 seconds, what was the magnitude of the average acceleration experienced by the ball during the time of contact?
### Multiple Choice Options
- ○ 220 m/s²
- ○ 390 m/s²
- ○ 41 m/s²
- ○ 170 m/s²
### Solution Steps
To find the magnitude of the average acceleration, use the formula:
\[ a = \frac{\Delta v}{\Delta t} \]
where \( \Delta v \) is the change in velocity and \( \Delta t \) is the time duration.
Let's calculate it step by step:
1. **Initial velocity (u)**: 37 m/s (as the ball was pitched towards the batter)
2. **Final velocity (v)**: 48 m/s (as the ball returns on the same line)
3. **Time of contact (t)**: 0.22 seconds
Change in velocity (\( \Delta v \)) is given by:
\[ \Delta v = v - (-u) = 48 - (-37) = 48 + 37 = 85 \, \text{m/s} \]
Average acceleration (\( a \)) is:
\[ a = \frac{85 \, \text{m/s}}{0.22 \, \text{s}} \approx 386.36 \, \text{m/s}^2 \]
Therefore, the closest correct option is \( 390 \, \text{m/s}^2 \).
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