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Math Help - Velocity/Acceleration Problem Help

  1. #1
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    Velocity/Acceleration Problem Help

    "A particle moves along the x-axis so that its velocity at time T is given by:

    v(T)= -(T+1)sin(T^2/2)

    At T=0 the particle is at position x=1.

    a) Find the acceleration of the particle at time T=2. Is the speed of the particle increasing at T=2? Why or why not?

    b) Find all times T in the open interval 0<T<3 when the particle changes direction. Justify your answer. "


    So I know to get the acceleration you have to derive the velocity function. This derivation consists of product quotient and chain rule.

    So I got

    V1(x)= Sin(t^2/2)*(-1)-(t+1)(cos(t^2/2))*(2*2t/4)

    Then you would just substitute in 2 for t to find the acceleration at t=2 right? Is this the correct derivative of the velocity function, and how would you go about part b?

    Thank you!
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  2. #2
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    Quote Originally Posted by roxy6 View Post
    "A particle moves along the x-axis so that its velocity at time T is given by:

    v(T)= -(T+1)sin(T^2/2)

    At T=0 the particle is at position x=1.

    a) Find the acceleration of the particle at time T=2. Is the speed of the particle increasing at T=2? Why or why not?

    b) Find all times T in the open interval 0<T<3 when the particle changes direction. Justify your answer. "


    So I know to get the acceleration you have to derive the velocity function. This derivation consists of product quotient and chain rule.

    So I got

    V1(x)= Sin(t^2/2)*(-1)-(t+1)(cos(t^2/2))*(2*2t/4)

    Then you would just substitute in 2 for t to find the acceleration at t=2 right? Is this the correct derivative of the velocity function, and how would you go about part b?

    Thank you!
    v(t) = -(t+1)\sin\left(\frac{t^2}{2}\right)

    a(t) = -(t+1)\cos\left(\frac{t^2}{2}\right) \cdot t - \sin\left(\frac{t^2}{2}\right)

    calculate a(2)

    for the (b) part, set v(t) = 0 and solve for t in the given interval. check the velocity on both sides of your solution to make sure it changes sign.
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  3. #3
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    Ok so for part A I did:

    a(2)= -(2+1)cos(2^2/2)*2-sin(2^2/2)

    a(2)= -4.083.


    The speed of the particle is decreasing because the acceleration is negative. Correct?

    And for part B I did:

    V(t) -(t+1)sin(t^2/2)=0
    V(0)= -(0+1)sin(0^2/2)=0
    V(1)= -(1+1)sin(1^2/2)= -1.438
    V(2) = -(2+2)sin(2^2/2)= -3.637
    V(3)= -(3+2)sin(3^2/2) = 4.887

    The particle changes direction when t=3 because the sign goes from negative to positive.


    Are these answers correct?
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  4. #4
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    Quote Originally Posted by roxy6 View Post
    Ok so for part A I did:

    a(2)= -(2+1)cos(2^2/2)*2-sin(2^2/2)

    a(2)= -4.083. no


    The speed of the particle is decreasing because the acceleration is negative. Correct? half correct ... your calculation and reasoning is wrong.

    And for part B I did:

    V(t) -(t+1)sin(t^2/2)=0
    V(0)= -(0+1)sin(0^2/2)=0
    V(1)= -(1+1)sin(1^2/2)= -1.438
    V(2) = -(2+2)sin(2^2/2)= -3.637
    V(3)= -(3+2)sin(3^2/2) = 4.887


    The particle changes direction when t=3 because the sign goes from negative to positive.

    no ... the particle changes direction when \textcolor{red}{\frac{t^2}{2} = \pi}

    ...
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  5. #5
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    I recalculated and got a(2)= 1.5875

    So the particle is speeding up because the sign is positive right?


    I am having a problem with part b and have been trying it for 2 hours now. Could you show me how to do it step by step?
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  6. #6
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    Quote Originally Posted by roxy6 View Post
    I recalculated and got a(2)= 1.5875

    So the particle is speeding up because the sign is positive right? no, again.

    the particle is slowing down because velocity and acceleration have opposite signs at t = 2. a(2) > 0 , v(2) < 0


    I am having a problem with part b and have been trying it for 2 hours now. Could you show me how to do it step by step?
    set v(t) = 0 ...

    -(t+1)\sin\left(\frac{t^2}{2}\right) = 0

    set each factor equal to 0 ...

    t+1 = 0 at t = -1, but that value of t is not in the interval 0 < t < 3

    \sin\left(\frac{t^2}{2}\right) = 0

    \frac{t^2}{2} = 0 , \pi , 2\pi , ...

    the only value that will get t in the desired interval 0 < t < 3 is

    \frac{t^2}{2} = \pi

    t^2 = 2\pi

    t = \sqrt{2\pi} \approx 2.507

    and the velocity changes sign there also.
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