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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?

College Physics
College Physics
11th Edition
ISBN:9781305952300
Author:Raymond A. Serway, Chris Vuille
Publisher:Raymond A. Serway, Chris Vuille
Chapter1: Units, Trigonometry. And Vectors
Section: Chapter Questions
Problem 1CQ: Estimate the order of magnitude of the length, in meters, of each of the following; (a) a mouse, (b)...
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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 \( 390 \, \text{m/s}^2 \).
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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