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Math Help - Position, Velocity, Acceleration problem

  1. #1
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    Position, Velocity, Acceleration problem

    A particle moves along the x-axis so that at any time t \geq 1 its accelaration is given by a(t)= \frac{1}{t}. At t=1, the velocity of the particle is v(1)=-2 and its position is x(1)=4.
    a) Find the velocity v(t) for t \geq 1.
    b) Find the position x(t) for t \geq 1.
    c) What is the position of the particle when it is farthest left?

    I answered parts (a) and (b) (unsure if they're correct):
    a) v(t)=\ln {t}-2
    b) x(t)=t\ln {t}-3t

    However I cannot answer part (c), here's what i have done(I have a bad feeling I am totally wrong):
    \int_1^\infty (\ln {t}-2) \, dt = \infty

    Can anyone help?
    Thanks in advance!
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  2. #2
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    Quote Originally Posted by xxsteelxx View Post
    A particle moves along the x-axis so that at any time t \geq 1 its accelaration is given by a(t)= \frac{1}{t}. At t=1, the velocity of the particle is v(1)=-2 and its position is x(1)=4.
    a) Find the velocity v(t) for t \geq 1.
    b) Find the position x(t) for t \geq 1.
    c) What is the position of the particle when it is farthest left?

    I answered parts (a) and (b) (unsure if they're correct):
    a) v(t)=\ln {t}-2
    b) x(t)=t\ln {t}-3t

    However I cannot answer part (c), here's what i have done(I have a bad feeling I am totally wrong):
    \int_1^\infty (\ln {t}-2) \, dt = \infty

    Can anyone help?
    Thanks in advance!
    a) is correct
    b) I think you are missing a constant

    for the last part here is a hint when does the velocity (part a) change from negative( going left) to positive (going right).
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  3. #3
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    Quote Originally Posted by xxsteelxx View Post
    A particle moves along the x-axis so that at any time t \geq 1 its accelaration is given by a(t)= \frac{1}{t}. At t=1, the velocity of the particle is v(1)=-2 and its position is x(1)=4.
    a) Find the velocity v(t) for t \geq 1.
    b) Find the position x(t) for t \geq 1.
    c) What is the position of the particle when it is farthest left?

    I answered parts (a) and (b) (unsure if they're correct):
    a) v(t)=\ln {t}-2
    b) x(t)=t\ln {t}-3t
    Is that "3" a typo? the integral of ln(t)- 2 is t ln(t)- 2t+ C. For x(1)= 4, you must have C= 6.

    However I cannot answer part (c), here's what i have done(I have a bad feeling I am totally wrong):
    \int_1^\infty (\ln {t}-2) \, dt = \infty

    Can anyone help?
    Thanks in advance!
    You find max and min of a function by setting the derivative equal to 0. Of course, here, the derivative is the speed. The particle will be farthest left when the speed was negative, becomes 0, and then becomes positive. When is ln(t)- 2= 0?
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  4. #4
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    Thank you very much guys! I understand part (c) now. However, I keep getting x(t)= t ln(t) - 3t +7 for part (b) ==> part c is therefore x(e^2)= 2e^2 - 3e^2 +7.
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