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Math Help - Applications of Derivatives

  1. #1
    Junior Member ImaCowOK's Avatar
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    Applications of Derivatives

    "A stone is thrown straight up from the edge of a roof, 725 feet above the ground, at a speed of 10 feet per second."

    "A. Remembering that the acceleration due to gravity is -32 feet per second squared, how high is the stone 2 seconds later?"

    I don't fully understand how to do these problems, the basic idea and steps. I have:

    dv/dt = -32 I think I have to take the antiderivative of this which would be:

    v = -32t + C

    I'm not sure what to do afterwards because I don't understand the basic steps behind solving these types of problems. I'm trying to follow the textbook, but it's confusing. There's more parts to the question after this, but I can't study because I don't understand.
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  2. #2
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    v = -32t + C.

    When t = 0, v = 10. So

    v = -32t + 10.

    Now v = dx/dt

    dx = -32t*dt + 10*dt

    Integrate it. When t = 0, x = 725 feet. Then solve for x at t = 2 s.
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  3. #3
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    Quote Originally Posted by ImaCowOK View Post
    "A stone is thrown straight up from the edge of a roof, 725 feet above the ground, at a speed of 10 feet per second."

    "A. Remembering that the acceleration due to gravity is -32 feet per second squared, how high is the stone 2 seconds later?"

    I don't fully understand how to do these problems, the basic idea and steps. I have:

    dv/dt = -32 I think I have to take the antiderivative of this which would be:

    v = -32t + C

    I'm not sure what to do afterwards because I don't understand the basic steps behind solving these types of problems. I'm trying to follow the textbook, but it's confusing. There's more parts to the question after this, but I can't study because I don't understand.
    Dear ImaCowOK,

    Apply the boundry conditions,(when t=0, v=10) then you will be able to find C. Since we want to find ditance the stone has travelled 2 seconds later, we have to get an equation which gives the distance in terms of the time. For that integrate the equation with respect to t again after substituting C. Keep in mind that v=\frac{ds}{dt} where v is the velocity and s is the displacement.

    Hope you will be able to solve this problem.
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