Mary walked north from her home to Sheila's home, which is 4.0 kilometers away. Then she turned right and walked another 3.0 kilometers to the supermarket, which is 5.0 kilometers from her own home. She walked the total distance in 1.5 hours. What were her average speed and average velocity?

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Mary walked north from her home to Sheila's home, which is 4.0 kilometers away. Then she turned right and walked another 3.0 kilometers to the supermarket, which is 5.0 kilometers from her own home. She walked the total distance in 1.5 hours. What were her average speed and average velocity?

Physics
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At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

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|dw:1434393199051:dw| the total distance wlaked is 4km + 3km = 7km she made it in 1.5hours, you can get the average speed using: \[s=\frac{d}{t}\] with: \(d\):the total distance \(t\):the total time find that first, then think what the average velocity should be :)
average speed=(4+3)Km/1.5 hours which is approx. 4.7Km/hour average velocity=5Km/1.5hours which is approx 3.3 Km/hour the direction of the average velocity is worked out like this: |dw:1434541251721:dw| sin theta=3/5 therefore theta= approximately 37 degrees so average velocity is 3.3Km/hour 37 degrees from North *Remember velocity is a vector quantity so we must always give direction. :) hope this helps!

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