At 7 am Joe starts sprinting at 6 miles per hour. At 7:10 am, Ken starts off after him. How fast must Ken run in order to overtake him at 7:30 am?

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At 7 am Joe starts sprinting at 6 miles per hour. At 7:10 am, Ken starts off after him. How fast must Ken run in order to overtake him at 7:30 am?

Mathematics
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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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9 mph 1/3*r = 1/2*6 r = 9
hence for joe: 6t=d (where t is time, and d is distance) for Ken: v=d/(20/60) no we know that after 30 minutes that's when ken is supposed to catch up that joe has already ran a distance of: 6*0.5=d, d=3miles substitute that back into v=d(3) then, speed of ken should be =3*3= 9 miles an hour

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