Jogging Jake runs along a train flatcar that moves at the

Conceptual Physics | 12th Edition | ISBN: 9780321909107 | Authors: Paul G. Hewitt

Problem 1R Chapter 3

Conceptual Physics | 12th Edition

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Conceptual Physics | 12th Edition | ISBN: 9780321909107 | Authors: Paul G. Hewitt

Conceptual Physics | 12th Edition

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Problem 1R

Jogging Jake runs along a train flatcar that moves at the velocities shown in positions A–D From greatest to least, rank the velocity of Jake relative to a stationary observer on the ground. (Call the direction to the right positive.)

Step-by-Step Solution:

Solution 1R Introduction We will use the concept of addition of relative velocity to find out the relative velocity of Jogging Jack in each cases and then we will compare. Step 1 If vtis the velocity of the train with respect to some stationary observer, and v js the velocity Jogging Jake with respect to train, then the velocity of the Jogging Jake with respect to the stationary observer will be v = v + v t j Step 2 (A) Here v t 8 m/s and v =j4 m/s Hence the relative velocity of Jack with respect to stationary observer is v = (8 m/s) + ( 4 m/s) = 4 m/s (B) In this case vt= 10 m/s and v = j m/s Hence v = ( 10 m/s) + (6 m/s) = 4 m/s (c) Here we have v = 6 m/s and v = 4 m/s, t j Hence we have v = (6 m/s) + (4 m/s) = 10 m/s (D) Here we have v =t 6m/s and v = 18jm/s Hence we have v = ( 6 m/s) + (18 m/s) = 12 m/s

Step 3 of 3

Chapter 3, Problem 1R is Solved
Textbook: Conceptual Physics
Edition: 12
Author: Paul G. Hewitt
ISBN: 9780321909107

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