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Math Help - Hot-Air Balloon

  1. #1
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    Hot-Air Balloon

    A hot-air balloon, headed due east at an average speed of 15 mph and at a constant altitude of 100 feet, passes over an intersection. Find an expression for its distance d (in feet) from the intersection t seconds later.

    Do I use D = rt?

    If so, then D = 100 feet, right?

    Then r = 15 mph, right?

    I got:

    100 = 15(t)

    100/15 = t

    Am I right or wrong?

    If wrong, what did I do wrong?
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  2. #2
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    Quote Originally Posted by symmetry View Post
    A hot-air balloon, headed due east at an average speed of 15 mph and at a constant altitude of 100 feet, passes over an intersection. Find an expression for its distance d (in feet) from the intersection t seconds later.

    Do I use D = rt?

    If so, then D = 100 feet, right?

    Then r = 15 mph, right?

    I got:

    100 = 15(t)

    100/15 = t

    Am I right or wrong?

    If wrong, what did I do wrong?
    x = rt is the balloon's horizontal distance from the intersection at t seconds.

    So x = 15t.

    BUT t is in seconds and the 15 is mi/h, so we need to change the unit. In addition we will want x to be in feet, so we need to change miles to feet. There are 3600 s in 1 h, and there are 5280 feet in 1 mile:
    \frac{15 \, mi}{1 \, hr} \cdot \frac{1 \, hr}{3600 \, s} \cdot \frac{5280 \, ft}{1 \, mi} = 22 ft/s

    So x = 22t in the units we want.

    Now, the balloon is at an altitude of 100 ft, so by the Pythagorean theorem the balloon is a distance of \sqrt{x^2 + (100)^2} feet from the intersection.
    d = \sqrt{(22t)^2 + (100)^2} = \sqrt{484t^2 + 10000} feet.

    -Dan
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  3. #3
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    ok

    I thank you for going the extra mile with this question.

    I don't think the question asked for an answer but it's good to know that an expression and answer were displayed by you here.

    Thanks.
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