An airplane, flying horizontally at an altitude of 1km.,passes directly over an observer. If the constant speed of the plane is 240kph, how fast is its distance from the observer increasing 30 seconds later?

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An airplane, flying horizontally at an altitude of 1km.,passes directly over an observer. If the constant speed of the plane is 240kph, how fast is its distance from the observer increasing 30 seconds later?

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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If we let t=0 when the plane passes over, the distance from the observer at time t will be Sqrt[1*1 + (240*t)^2] where t is in hours (via Pythagorean theorem). The change in distance is the derivative of this, or 57600t/Sqrt[1+57600*t^2]. 30 seconds means t=1/120 (since we're measuring t in hours), so the change in distance over time is 96*Sqrt[5] kph or about 214.66 kph.
wow thanks a lot
ooops i didn't get the same answer

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Other answers:

isn't 2*57600t/Sqrt[1+57600*t^2] ???

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