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Math Help - Please Please Help !!!!

  1. #1
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    Please Please Help !!!!

    A particle moves along the x-axis so that its velocity is given by
    v(t)= -(t+1)sin(t^2/2)
    ****at time t=0, the particle is at position x=1

    a) Find the acceleation 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.

    c)Find the total distance traveled by the particle from time t=0 until time t=3

    d) During the time interval 0<=t<=3, , what is the greatest distance between the particle and the origin?

    I WOULD GREATLY APPRECIATE IT IF SOMEONE WOULD HELP ME OUT ON THIS PROBLEM AND GIVE ME THE CORRECT ANSWERS SO I CAN SEE IF MINE ARE CORRECT. I'M DESPERATE , PLEASE HELP.
    - ALLEN
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  2. #2
    Grand Panjandrum
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    Quote Originally Posted by sheven
    A particle moves along the x-axis so that its velocity is given by
    v(t)= -(t+1)sin(t^2/2)
    ****at time t=0, the particle is at position x=1

    a) Find the acceleation of the particle at time t=2. Is the speed of the particle increasing at t=2? Why or Why not?
    a)

    Acceleration is the rate of change of speed, so:

    <br />
a(t)=\frac{dv}{dt}=-\sin(t^2/2)-(t+1)t\cos(t^2/2)<br />

    where the derivative is found using a combination of the product rule and
    the chain rule.

    Substituting t=2 into this gives:

    a(2)\approx 1.588

    This is positive, and positive acceleration is the same thing as increasing
    speed, so the speed is increasing.

    RonL
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  3. #3
    Grand Panjandrum
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    Quote Originally Posted by sheven
    A particle moves along the x-axis so that its velocity is given by
    v(t)= -(t+1)sin(t^2/2)
    ****at time t=0, the particle is at position x=1
    b) Find all times t in the open interval 0,t,3 when the particle changes direction.Justify your answer.
    b)

    When the particle changes direction the velocity changes sign, and as
    the velocity is a continuous function of time it must be zero at those
    times that it changes direction.

    If the velocity is zeros then:

    <br />
v(t)=-(t+1)\sin(t^2/2)=0<br />

    For t \in (0,3) (we ignore the end points as you can't change direction
    at these points) this only occurs at the zeros of \sin(t^2/2).
    These are t=\sqrt{2 n \pi},\ n=1,2, \dots, and we now need to check
    which values of n give t \in (0,3).

    The only value of n that gives a t \in (0,3) is 1 and the corresponding time
    is \approx 2.5066

    You will now have to check that this is in fact a point at which the velocity
    changes sign rather than a point at which it is zero but does not change sign.
    This means we need to check that the acceleration does not equal zero at
    this time. Plugging t=2.5066 into the equation for the acceleration
    found in part a) shows that the acceleration is indeed not zero, so the particle
    changes direction at this time.

    RonL

    RonL
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  4. #4
    Grand Panjandrum
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    Quote Originally Posted by sheven
    A particle moves along the x-axis so that its velocity is given by
    v(t)= -(t+1)sin(t^2/2)
    ****at time t=0, the particle is at position x=1
    c)Find the total distance traveled by the particle from time t=0 until time t=3
    c)

    I find this slightly ambiguous does this mean the total distance travelled by
    the partice or the distance of the particle at time t=3 from its position
    at time t=0?

    The position s(t) at time t is given by:

    <br />
s(t)=\int_0^t v(\xi) d\xi +s(0) =\int_0^t -(\xi+1)\sin(\xi^2/2) d\xi +s(0) <br />

    The problem here is that this last integral is non-elementary (it can be written
    as the sum of an elementary function and Fresnel's S function). This problem
    can be overcome by integrating the functions numericaly.

    Now the distance travelled by the particle between t=0 and t=3 is:

    <br />
\rm{dist.}=\int_0^3 |v(\xi)| d\xi +s(0) =\int_0^3 |(\xi+1)\sin(\xi^2/2)| d\xi\approx 4.33 <br />

    and the displacement of the final position from the starting position is:

    <br />
\rm{disp.}=\int_0^3 v(\xi) d\xi =\int_0^3 -(\xi+1)\sin(\xi^2/2) d\xi\approx -2.20 <br />


    d) During the time interval 0<=t<=3, , what is the greatest distance between the particle and the origin?
    Again this can be done numericaly and is \approx 2.27.

    RonL
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  5. #5
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    I am not quite sure how you got your answer for (d). Could you please elaborate your work on that problem please? Thanks
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  6. #6
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    Quote Originally Posted by Vigo
    I am not quite sure how you got your answer for (d). Could you please elaborate your work on that problem please? Thanks
    I take the form

    <br />
s(t)=\int_0^t -(\xi+1)\sin(\xi^2/2) d\xi +1<br />

    You want to find the point that it is furtherest form 0 on [ 0,3 ],
    so differentiate and set the derivative to zero. Now:

    <br />
\frac{d}{dt} s(t) =v(t)=-(t+1)\sin(t^2/2)<br />
,

    but we have already found the zeros of this in [0,3], so we
    need only check the end points and this zero at t\approx 2.5066
    to find the extrema of s(t).

    Note that at this point  s \approx -2.27 , but the question asks
    for the graetest distance so the sign is dropped.

    RonL
    Last edited by CaptainBlack; March 15th 2006 at 11:21 PM.
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