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The graph of acceleration (on the vertical axis) versus time (on the horizontal axis) is a horizontal line given by a, = 2.0 m/s? from 0 seconds to 4 seconds and very quickly changes to a horizontal line given by ax = -- 3.0 m/s2 from 4 seconds to 7 seconds. Find the change in velocity from time equals 1 second to time equals 5 seconds. O 8.0 m/s in the +x-direction. O 9.0 m/s in the +x-direction. O 3.0 m/s in the --x-direction. O 3.0 m/s in the +x-direction.

The graph of acceleration (on the vertical axis) versus time (on the horizontal axis) is a horizontal line given by a, = 2.0 m/s? from 0 seconds to 4 seconds and very quickly changes to a horizontal line given by ax = -- 3.0 m/s2 from 4 seconds to 7 seconds. Find the change in velocity from time equals 1 second to time equals 5 seconds. O 8.0 m/s in the +x-direction. O 9.0 m/s in the +x-direction. O 3.0 m/s in the --x-direction. O 3.0 m/s in the +x-direction.

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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**Acceleration and Velocity Change: Problem Analysis** **Problem Statement:** The graph of acceleration (on the vertical axis) versus time (on the horizontal axis) is a horizontal line given by \( a_x = 2.0 \, \mathrm{m/s^2} \) from 0 seconds to 4 seconds, and very quickly changes to a horizontal line given by \( a_x = -3.0 \, \mathrm{m/s^2} \) from 4 seconds to 7 seconds. Find the change in velocity from time equals 1 second to time equals 5 seconds. **Given:** - \( a_x = 2.0 \, \mathrm{m/s^2} \) from \( t = 0 \) seconds to \( t = 4 \) seconds. - \( a_x = -3.0 \, \mathrm{m/s^2} \) from \( t = 4 \) seconds to \( t = 7 \) seconds. **Objective:** Calculate the change in velocity (\( \Delta v \)) from \( t = 1 \) second to \( t = 5 \) seconds. **Multiple Choice Answers:** - \( \bigcirc \) 8.0 m/s in the +x-direction. - \( \bigcirc \) 9.0 m/s in the +x-direction. - \( \bigcirc \) 3.0 m/s in the -x-direction. - \( \bigcirc \) 3.0 m/s in the +x-direction. **Solution Steps:** 1. **Calculate the change in velocity from \( t = 1 \) second to \( t = 4 \) seconds:** - From \( 1 \) to \( 4 \) seconds, the acceleration (\( a_x \)) is \( 2.0 \, \mathrm{m/s^2} \). - Time interval (\( \Delta t \)) is \( 4 - 1 = 3 \) seconds. - Change in velocity (\( \Delta v_1 \)) is given by \( a_x \times \Delta t \): \[ \Delta v_1 = 2.0 \, \mathrm{m/s^2} \times 3 \, \mathrm{s} = 6.0 \, \mathrm{m/s}
Transcribed Image Text:**Acceleration and Velocity Change: Problem Analysis** **Problem Statement:** The graph of acceleration (on the vertical axis) versus time (on the horizontal axis) is a horizontal line given by \( a_x = 2.0 \, \mathrm{m/s^2} \) from 0 seconds to 4 seconds, and very quickly changes to a horizontal line given by \( a_x = -3.0 \, \mathrm{m/s^2} \) from 4 seconds to 7 seconds. Find the change in velocity from time equals 1 second to time equals 5 seconds. **Given:** - \( a_x = 2.0 \, \mathrm{m/s^2} \) from \( t = 0 \) seconds to \( t = 4 \) seconds. - \( a_x = -3.0 \, \mathrm{m/s^2} \) from \( t = 4 \) seconds to \( t = 7 \) seconds. **Objective:** Calculate the change in velocity (\( \Delta v \)) from \( t = 1 \) second to \( t = 5 \) seconds. **Multiple Choice Answers:** - \( \bigcirc \) 8.0 m/s in the +x-direction. - \( \bigcirc \) 9.0 m/s in the +x-direction. - \( \bigcirc \) 3.0 m/s in the -x-direction. - \( \bigcirc \) 3.0 m/s in the +x-direction. **Solution Steps:** 1. **Calculate the change in velocity from \( t = 1 \) second to \( t = 4 \) seconds:** - From \( 1 \) to \( 4 \) seconds, the acceleration (\( a_x \)) is \( 2.0 \, \mathrm{m/s^2} \). - Time interval (\( \Delta t \)) is \( 4 - 1 = 3 \) seconds. - Change in velocity (\( \Delta v_1 \)) is given by \( a_x \times \Delta t \): \[ \Delta v_1 = 2.0 \, \mathrm{m/s^2} \times 3 \, \mathrm{s} = 6.0 \, \mathrm{m/s}
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