A dog running in an open field has components of velocity

University Physics | 13th Edition | ISBN: 9780321675460 | Authors: Hugh D. Young, Roger A. Freedman

Problem 6E Chapter 3

University Physics | 13th Edition

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University Physics | 13th Edition | ISBN: 9780321675460 | Authors: Hugh D. Young, Roger A. Freedman

University Physics | 13th Edition

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Problem 6E

A dog running in an open field has components of velocity vx = 2.6 m/s and vy = -1.8 m/s at t1 = 10.0 s. For the time interval from t1 = 10.0 s to t2 = 20.0 s, the average acceleration of the dog has magnitude 0.45 m/s2 and direction 31.0o measured from the + x-axis toward the + y-axis. At t2 = 20.0 s, (a) what are the x- and y-components of the dog’s velocity? (b) What are the magnitude and direction of the dog’s velocity? (c) Sketch the velocity vectors at t1 and t2. How do these two vectors differ?

Step-by-Step Solution:

Solution 6E Step 1 (a) Initially at 1 = 10.0 sthe x component of the velocity is v = 2x6 m/s, and y component of the velocity is v y 1.8 m/s. In the interval between t =110.0 s to t = 20.0 s the magnitude of the average acceleration is |a| = 0.45 m/s and the angle is = 31.0°. 2 2 So in this interval the x component is a = xa|cos = (0.45 m/s )cos31.0° = 0.39 m/s And the y component of the acceleration is a = |a|sin = (0.45 m/s )sin31.0° = 0.23 m/s . 2 y And in this interval of time the change in time is t = t 2 = 10 s 10 s = 10 s. So the x component of the velocity at at t = 20 s is 2 v x2= v x a x = (2.6 m/s) + (0.39 m/s )(10 s) = 6.5 m/s And the y component of the velocity is 2 v y2= v y a y = ( 1.8 m/s) + (0.23 m/s )(10.0 s) = 0.50 m/s Step 2 (b) The magnitude of velocity is |v| = v 2 + v 2 = 6.51 m/s x2 y2 And the direction of the velocity is = tan (1 vy) = tan (1 0.50 m/) = 4.4° anticlockwise from the positive vx2 6.5 m/s x axis.

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Chapter 3, Problem 6E is Solved
Textbook: University Physics
Edition: 13
Author: Hugh D. Young, Roger A. Freedman
ISBN: 9780321675460

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