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Math Help - traveling car

  1. #1
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    please help its urgent

    A car traveling at a rate of 30 ft/sec is approaching an intersection. When the car is 120 ft from the intersection, a truck traveling at a rate of 40 ft/sec crosses the intersection. If the roads are at right angles to each other and the truck leaves the intersection after 2 seconds, how fast are the car and the truck separating when the truck leaves the intersection?
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  2. #2
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    " A car traveling at a rate of 30 ft/sec is approaching an intersection. When the car is 120 ft from the intersection, a truck traveling at a rate of 40 ft/sec crosses the intersection. If the roads are at right angles to each other and the truck leaves the intersection after 2 seconds, how fast are the car and the truck separating when the truck leaves the intersection? "

    Here is one way.

    After some time t seconds:
    >>>the car travelled 30*t ft, so it is now (120 -30t) ft from the intersection.
    >>>the truck travelled 40*t ft, so it is now 40t ft from the intersection.

    Let x = distance between the car and truck at any time, in ft.

    Imagine the figure,
    It is a right triangle, with these:
    hypotenuse = x
    one leg = 120 -30t
    the other leg = 40t
    So, by Pythagorean theorem,
    x^2 = (120 -30t)^2 +(40t)^2 ----(i)
    Differentiate both sides of (i) with respect to time t,
    2x(dx/dt) = 2(120 -30t)(-30) +2(40t)(40)
    2x(dx/dt) = -7200 +1800t +3200t
    2x(dx/dt) = 5000t -7200 ---------------(ii)

    "...and the truck leaves the intersection after 2 seconds, how fast are the car and the truck separating when the truck leaves the intersection?"

    That means, when t=2sec, what is dx/dt?

    In (ii), we know t=2, but we don't now x.

    In (i) we can get x when t=2, so,
    x^2 = (120 -30t)^2 +(40t)^2 ----(i)
    x^2 = (120 -30*2)^2 +(40*2)^2
    x^2 = 60^2 +80^2
    [if you know the 3-4-5 right triangle, you can say here now that x=100.]
    x^2 = 3600 +6400 = 10000
    x = sqrt(10000) = 100 ft

    Substitute those into (ii),
    2x(dx/dt) = 5000t -7200 ---------------(ii)
    2(100)(dx/dt) = 5000(2) -7200
    200(dx/dt) = 2800
    dx/dt = 2800/200 = 14 ft/sec ------***

    Therefore, at that time, the car and truck are separating at the rate of 14 ft/sec. ---answer.
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  3. #3
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    thankyou

    thankyou very much
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