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(II) A dog runs 120 $\mathrm{m}$ away from its master in a straight linein 8.4 $\mathrm{s}$ , and then runs halfway back in one-third the time.Calculate $(a)$ its average speed and $(b)$ its average velocity.

a) 16 m/sb) $5 m / s,$ away from owner

Physics 101 Mechanics

Chapter 2

Describing Motion: Kinematics in One Dimension

Physics Basics

Motion Along a Straight Line

Motion in 2d or 3d

Newton's Laws of Motion

Cornell University

University of Washington

Hope College

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So here we can say that we want to find rather for part of the average speed for part B the average velocity so we can say that the distance traveled would be 120 meters plus 1/2 of 120 meters, giving us 180 meters now, however, are displacement from the origin would be 120 meters away, minus 1/2 because we're going back where it's actor traveling 120 meters and this is equaling 60 meters. Ah, we can say our time total would be equal to 8.4 seconds off, plus 1/2 times a father 1/3 my apologies times 8.4 seconds, giving us 11 0.2 seconds and now for part A to find the average per day to find the average speed. This would be equal to the distance divided by time, so this would be equal to 180 meters, divided by 11.2 seconds, giving us 16 approximately 16 meters per second. Now for part B, we want the average velocity and here velocity on Lee depends on this on the displacement. So not the total distance, but only the displacement. So here we can say that the velocity average would be equal to 60 meters divided by 11.2 seconds. And this is giving us five approximately five meters per second. Again, we have to around two. Um, here we have two round toe, one significant figure because 60 meters on Lee has one significant figure here. We can round 22 significant figures because 180 meters has two significant figures. Ah, and then we can say here this would be positive because this would be in original direction, direction. It's the end of the solution. Thank you for watching.

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