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Math Help - Mechanics - Trick question?

  1. #1
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    Mechanics - Trick question?

    Hey all i was doing this mechanics 1 question but two parts of it dont make sense to me so i was wondering if i was doing the question correct.



    Find velocity of the particle,  \dot{x}(t) for  t > 0

     \ddot{x} = \sin t. Integrating this gives
    \dot{x} = - \cos t + c

    Now thats the bit that tricks me, its says for t > 0. But we are not given any conditions for t > 0. So do i just plug in where
    t = 0, \dot{x}=0 hence c = 0?
    <br />
\therefore \dot{x} = - \cos t

    The same problem when finding the position of the particle for t>0.

    Do i just integrate \dot{x} and sub in where t=0, x=0 ?


    Finally, i dont have a clue on how to draw the distance or the velocity.

    Hopefully you lot can help me out, .
    Thanks.
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  2. #2
    Senior Member AllanCuz's Avatar
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    Quote Originally Posted by simpleas123 View Post
    Hey all i was doing this mechanics 1 question but two parts of it dont make sense to me so i was wondering if i was doing the question correct.



    Find velocity of the particle,  \dot{x}(t) for  t > 0

     \ddot{x} = \sin t. Integrating this gives
    \dot{x} = - \cos t + c

    Now thats the bit that tricks me, its says for t > 0. But we are not given any conditions for t > 0. So do i just plug in where t = 0, \dot{x}=0 hence c = 0?
    <br />
\therefore \dot{x} = - \cos t

    The same problem when finding the position of the particle for t>0.

    Do i just integrate \dot{x} and sub in where t=0, x=0 ?


    Finally, i dont have a clue on how to draw the distance or the velocity.

    Hopefully you lot can help me out, .
    Thanks.
    You found the velocity required for question 1,

     \frac{dx}{dy} = -cost + c for  t > 0

    This is the velocity for all the points greater then t. It doesn't ask for any more then this, and if you notice, in the second question it asks us to graph the above function. That is how you'll know what the values will be.

    Note though that,

     y^{ \prime } (0)= 0 = -1 + c so  c = 1 and

     y^{ \prime } = -cost + 1

    For question 2,

    We need to graph  y^{ \prime } = -cost + 1

    You can either do this via tech (i.e. tx calculator or an online grapher) or you can notice that this graph is really just a translated version of  cost

    You know how to draw  cost right? So draw that on paper for the given period, then draw the negative of that because we have  - cost then after doing that, add 1 to account for our constant. This will get you the graph,

    For Question 3,

     \frac{dx}{dy} = -cost + 1

     x(t) = distance = -sint + x + c

    Sketch this in the same way as the other one
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  3. #3
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    Quote Originally Posted by AllanCuz View Post
    You found the velocity required for question 1,

     \frac{dx}{dy} = -cost + c for  t > 0

    This is the velocity for all the points greater then t. It doesn't ask for any more then this, and if you notice, in the second question it asks us to graph the above function. That is how you'll know what the values will be.

    Note though that,

     y^{ \prime } (0)= 0 = -1 + c so  c = 1 and

     y^{ \prime } = -cost + 1

    For question 2,

    We need to graph  y^{ \prime } = -cost + 1

    You can either do this via tech (i.e. tx calculator or an online grapher) or you can notice that this graph is really just a translated version of  cost

    You know how to draw  cost right? So draw that on paper for the given period, then draw the negative of that because we have  - cost then after doing that, add 1 to account for our constant. This will get you the graph,

    For Question 3,

     \frac{dx}{dy} = -cost + 1

     x(t) = distance = -sint + x + c

    Sketch this in the same way as the other one
     y^{ \prime } (0)= 0 = -1 + c so  c = 1 and

     y^{ \prime } = -cost + 1

    how did you derive that?
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  4. #4
    Senior Member AllanCuz's Avatar
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    Quote Originally Posted by simpleas123 View Post
     y^{ \prime } (0)= 0 = -1 + c so  c = 1 and

     y^{ \prime } = -cost + 1

    how did you derive that?
    I didn't derive it you did!

     \frac{ d^2 x }{dt^2} = sint

     d^2 x = sint dt^2

     \int d^2 x = \int sint dt^2

     dx = (-cost + c ) dt

     \frac{dx}{dt} = y^{ \prime }  = -cost + c

    The initial conditions in the question say  y^{ \prime } (0) = 0

    So,

     0 = -cos(0) + c

     c = 1
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  5. #5
    Grand Panjandrum
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    Quote Originally Posted by simpleas123 View Post
    Hey all i was doing this mechanics 1 question but two parts of it dont make sense to me so i was wondering if i was doing the question correct.



    Find velocity of the particle,  \dot{x}(t) for  t > 0

     \ddot{x} = \sin t. Integrating this gives
    \dot{x} = - \cos t + c

    Now thats the bit that tricks me, its says for t > 0. But we are not given any conditions for t > 0. So do i just plug in where
    t = 0, \dot{x}=0 hence c = 0?
    <br />
\therefore \dot{x} = - \cos t

    The same problem when finding the position of the particle for t>0.

    Do i just integrate \dot{x} and sub in where t=0, x=0 ?


    Finally, i dont have a clue on how to draw the distance or the velocity.

    Hopefully you lot can help me out, .
    Thanks.
    First the general solution to the homogeneous equation:

    \frac{d^2x}{dt^2}=0

    is

    x(t)=at+c

    Then a particular integral of:

    \frac{d^2x}{dt^2}=\sin(t)

    is:

    x(t)=-\sin(t)

    So the general solution of the original ODE is:

    x(t)=(at+c)-\sin(t)

    Applying the initial conditions gives:

    x(t)=t-\sin(t)

    which should allow all the questions to be answered.

    Off course you should get exactly the same answer by integrating twice since the left hand side is just the second time derivative and the right a function of time only.

    CB
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