A turtle and a hare race. The hare takes off at 4.5 m/s for 15 minutes and the turtle takes off at 0.45 m/s. The hare then slows down to 1.5 m/s for 10 mins. Then it falls asleep for 20 minutes. It wakes up and sprints for the next 15 mins at 6.2 m/s. a. What is the hare's avg. velocity? b. What is the total displacement of the hare? c. What is the turtle's displacement if he continued moving at the same rate for the same amount of time as the hare's total trip?

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A turtle and a hare race. The hare takes off at 4.5 m/s for 15 minutes and the turtle takes off at 0.45 m/s. The hare then slows down to 1.5 m/s for 10 mins. Then it falls asleep for 20 minutes. It wakes up and sprints for the next 15 mins at 6.2 m/s. a. What is the hare's avg. velocity? b. What is the total displacement of the hare? c. What is the turtle's displacement if he continued moving at the same rate for the same amount of time as the hare's total trip?

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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Can you tell me whats the duration for which Hare ran with 4.5m/s????
15 minutes, sorry.
\[v _{avg}=\frac{ Total Displacement }{Total Time }\] So, \[v _{avg}=2.925m/s\]for Hare Total displacement=10530m {Vel.xTime(In sec)} \[Turtle's Displacement=Vel. \times Time=0.45 \times 3600= 16202m\]

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