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Thread: Applied Calc Problem

  1. #1
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    Applied Calc Problem

    Define s(t)= 2sin(t) + sqrt(2t) to be the position of a particle after time t in seconds. Bound t by the domain 0 is less than or equal to t, t is less than or equal to 2pi.
    a. Find the velocity and acceleration functions of s(t).
    b. Find when the particle is at rest.
    c. Find the times when the particle is moving forward and moving backward from its initial position.

    A. I got s'(t)=v(t)= 2cos(t) + 1/sqrt(t)
    and s''(t)=a(t)= -2sin(t) + 1/(2)sqrt(t^3)
    Is this correct so far?

    B. I set v(t)=0, so: 2cos(t) + 1/sqrt(t) = 0
    How do I solve for cos(t)?

    C. I have no idea where to begin.
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  2. #2
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    Quote Originally Posted by snow2612 View Post
    Define s(t)= 2sin(t) + sqrt(2t) to be the position of a particle after time t in seconds. Bound t by the domain 0 is less than or equal to t, t is less than or equal to 2pi.
    a. Find the velocity and acceleration functions of s(t).
    b. Find when the particle is at rest.
    c. Find the times when the particle is moving forward and moving backward from its initial position.

    A. I got s'(t)=v(t)= 2cos(t) + 1/sqrt(t)
    and s''(t)=a(t)= -2sin(t) + 1/(2)sqrt(t^3)
    Is this correct so far?

    B. I set v(t)=0, so: 2cos(t) + 1/sqrt(t) = 0
    How do I solve for cos(t)?

    use technology (a calculator) to find the solutions ... graph v(t) in the given domain and locate the zeros.

    C. I have no idea where to begin.

    look at the graph of v(t) ... (+) values of v(t), the particle moves right.
    (-) values, the particle moves left.
    ...
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  3. #3
    Grand Panjandrum
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    Quote Originally Posted by snow2612 View Post
    Define s(t)= 2sin(t) + sqrt(2t) to be the position of a particle after time t in seconds. Bound t by the domain 0 is less than or equal to t, t is less than or equal to 2pi.
    a. Find the velocity and acceleration functions of s(t).
    b. Find when the particle is at rest.
    c. Find the times when the particle is moving forward and moving backward from its initial position.

    A. I got s'(t)=v(t)= 2cos(t) + 1/sqrt(t)
    and s''(t)=a(t)= -2sin(t) + 1/(2)sqrt(t^3)
    Is this correct so far?.
    s'(t)=2\cos(t)+\frac{1}{\sqrt{2t}}

    CB
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