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Math Help - Two antiderivative problems!!!

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    Two antiderivative problems!!!

    1. Evaluate the indefinite integral.



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

    A. Remembering that the acceleration due to gravity is , how high is the stone seconds later?

    B. At what time does the stone hit the ground?

    C. What is the velocity of the stone when it hits the ground?
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    Quote Originally Posted by Kayla_N View Post
    1. Evaluate the indefinite integral.



    Let u=5+14x^4\Rightarrow du=56x^3\,dx.

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

    A. Remembering that the acceleration due to gravity is , how high is the stone seconds later?
    Integrate the acceleration function to get the velocity function v. You can solve for the constant of integration by noting that v(0)=20.

    Integrate v to get the stone's position function s. Set s(0)=775 and you can solve for the constant of integration. Then just evaluate s(3).

    B. At what time does the stone hit the ground?
    That is, for what value of t is s(t)=0?

    C. What is the velocity of the stone when it hits the ground?
    Evaluate the velocity function at the time found in part (B).
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    I still dont understand what you wrote for number 2..can you explain a little bit more.???
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    Quote Originally Posted by Kayla_N View Post
    I still dont understand what you wrote for number 2..can you explain a little bit more.???
    Velocity is the derivative of position; acceleration is the derivative of velocity.

    The stone is undergoing a constant -32\text{ ft./s}^2 acceleration due to gravity, so the acceleration function is

    a(t)=-32.

    The velocity function is therefore

    v(t)=\int a(t)\,dt=\int(-32)\,dt=-32t+C_0.

    We know that the initial velocity is 20 ft./s, so

    v(0)=20\Rightarrow -32\cdot0+C_0=20\Rightarrow C_0=20

    and

    v(t)=20-32t.

    Work similarly to find the position function. Can you continue?
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    Quote Originally Posted by Reckoner View Post
    Velocity is the derivative of position; acceleration is the derivative of velocity.

    The stone is undergoing a constant -32\text{ ft./s}^2 acceleration due to gravity, so the acceleration function is

    a(t)=-32.

    The velocity function is therefore

    v(t)=\int a(t)\,dt=\int(-32)\,dt=-32t+C_0.

    We know that the initial velocity is 20 ft./s, so

    v(0)=20\Rightarrow -32\cdot0+C_0=20\Rightarrow C_0=20

    and

    v(t)=20-32t.

    Work similarly to find the position function. Can you continue?

    since v(t)=20-32t, so t= .625.
    Then to find s(t)= \int (-32+20)dt
    s(t)=-16t^2+20t+C2
    3=-16(0)+20(0)+c2
    C2=3??
    Then i dont know what else to do!!!
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    Quote Originally Posted by Kayla_N View Post
    since v(t)=20-32t, so t= .625.
    At t=0.625, the velocity is zero (i.e., the stone is not moving). This will occur when the stone reaches its maximum height. Were you asked to find this?

    Then to find s(t)= \int (-32+20)dt
    s(t)=-16t^2+20t+C2
    3=-16(0)+20(0)+c2
    What are you doing? The initial height is 775 ft., not 3. After you solve for C_2, evaluate s(3) to get the position after 3 seconds.
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    So the s(t) -16t^2+20t+739 right???
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    Quote Originally Posted by Kayla_N View Post
    So the s(t) -16t^2+20t+739 right???
    Where did you get 739 from?
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    Quote Originally Posted by Reckoner View Post
    Where did you get 739 from?
    well 739 is wrong, i did again 775=-16t^2+20t+C
    then i plug in t=3 and i got C=859?

    Jeeze i'm so lost.
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    Quote Originally Posted by Kayla_N View Post
    well 739 is wrong, i did again 775=-16t^2+20t+C
    then i plug in t=3 and i got C=859?
    Think about what you are doing. The stone's initial height (at 0 seconds) is 775 ft., so when t=0,\;s(t)=775. So why are you setting t=3?
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    i'm trying to calculate the height at 3 seconds.. Can you please show me how to get part A??
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    Quote Originally Posted by Kayla_N View Post
    i'm trying to calculate the height at 3 seconds.. Can you please show me how to get part A??
    Do as I said:

    We have s(t)=-16t^2+20t+C_2, and s(0)=775 (as I mentioned above). Therefore,

    s(0)=-16\cdot0^2+20\cdot0+C_2=775

    \Rightarrow C_2=775

    and

    s(t)=-16t^2+20t+775.

    Now find s(3).
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    ok and i got 691. From there i went on to second part...s(t)=0
    0=-16t^2+20t+775. I used quadratic formala and got 7.6127 sec. then moved on to 3 part, i used V= (7.6127)(-32)+20= -223.6068.

    Thanks for your help...
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    Quote Originally Posted by Kayla_N View Post
    ok and i got 691. From there i went on to second part...s(t)=0
    0=-16t^2+20t+775. I used quadratic formala and got 7.6127 sec. then moved on to 3 part, i used V= (7.6127)(-32)+20= -223.6068.
    Looks good!
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    Thanks...just part A that i'm stuck on..Thanks again for you help. I really really appreciated.
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