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Math Help - Expressions

  1. #1
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    Expressions

    Hey in the below attachment is the answers for part
    i) <br />
a = 2x-4<br />
    ii) <br />
x = -\frac{1}{t}\ +2<br />

    and how would i do part iii) thanks again Nath
    Attached Thumbnails Attached Thumbnails Expressions-particle.jpg  
    Last edited by CaptainBlack; July 10th 2006 at 12:14 AM.
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  2. #2
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    Quote Originally Posted by nath_quam
    Hey in the below attachment is the answers for part
    i) <br />
a = 2x-4<br />
    ii) <br />
x = -\frac{1}{t}\ +2<br />

    and how would i do part iii) thanks again Nath
    You are given:

    v=(2-x)^2,

    so for part (i):

    <br />
a=\frac{dv}{dt}=\frac{d}{dt}(2-x)^2=2(2-x)(-1)\frac{dx}{dt}=-2(2-x)^3<br />

    If as pointed out by nath_quam v is in cm/s but acceleration is wanted in
    m/s^2, in the above equation the acceleration is in cm/s^2 and
    needs to be divided by 100 to convert it into m/s^2. However as Topsquark
    points out there is probably a typo in the question and what appears as
    "c m/s", should read "v m/s" then the dimensions in the above are OK
    (unless someone else spots an error ).

    RonL
    Last edited by CaptainBlack; July 10th 2006 at 03:34 AM.
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  3. #3
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    Quote Originally Posted by nath_quam
    ii) <br />
x = -\frac{1}{t}\ +2<br />
    How does this satisfy the initial condition that at t=0, x=0\ \?

    RonL
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  4. #4
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    yerh i see looks like i have to try again how would i do ii and iii anyway thanks nath
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  5. #5
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    Quote Originally Posted by nath_quam
    yerh i see looks like i have to try again how would i do ii and iii anyway thanks nath
    Like this:

    <br />
v=\frac{dx}{dt}=(2-x)^2<br />

    which is a differential equation of variables seperable type, and so:

    <br />
\int\frac{dx}{(2-x)^2}=\int dt<br />
,

    which may be integrated to give:

    <br />
\frac{1}{2-x}=t+const<br />

    and as x=0 when t=0,\ const=1/2, then rearranging:

    x=\frac{4t}{2t+1}

    RonL
    Last edited by CaptainBlack; July 10th 2006 at 02:02 AM.
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  6. #6
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    Quote Originally Posted by nath_quam
    Hey in the below attachment is the answers for part
    i) <br />
a = 2x-4<br />
    ii) <br />
x = -\frac{1}{t}\ +2<br />

    and how would i do part iii) thanks again Nath
    iii) find the distance from O when when v=1.

    If v=1, then as v=(2-x)^2:

    <br />
1=(2-x)^2<br />
,

    or:

    <br />
x^2-4x+3=0<br />
,

    which can be solved using the quadratic formula, to give the xs
    corresponding to v=1. The required distance is |x|.

    There may be an implied requirement that the corresponding time be positive,
    so the times at which the partical is at these positions may need to be checked.

    RonL
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  7. #7
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    Quote Originally Posted by CaptainBlack
    iii) find the distance from O when when v=1.

    If v=1, then as v=(2-x)^2:

    <br />
1=(2-x)^2<br />
,

    or:

    <br />
x^2-4x+3=0<br />
,

    which can be solved using the quadratic formula, to give the xs
    corresponding to v=1. The required distance is |x|.

    There may be an implied requirement that the corresponding time be positive,
    so the times at which the partical is at these positions may need to be checked.

    RonL
    Two things with this part firstly the questions states velocity is in cm/s our question is in m/s and velocity isn't speed is it but distance over time
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  8. #8
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    Quote Originally Posted by nath_quam
    Two things with this part firstly the questions states velocity is in cm/s our question is in m/s and velocity isn't speed is it but distance over time
    I must say that I hadn't noticed the mixed units, but I'm sure you can sort
    that out from what I/we have done. (this is partialy a problem with having the
    statement of the problem/s in an attachment, a respondent cannot see what
    they are replying to on the same page as that on which they are typing the
    reply - I think in future I will have to retype the question so I can see it
    while replying - or maybe just open two browser windows one for the reply
    and another for the question)

    The second point in this case is irrelevant, as the velocity is always positive,
    and so is the same thing as the speed here since speed=|velocity| (we are
    talking instantaneous speeds and velocities not average speeds and velocities).

    RonL
    Last edited by CaptainBlack; July 10th 2006 at 08:52 AM.
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  9. #9
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    Quote Originally Posted by nath_quam
    Two things with this part firstly the questions states velocity is in cm/s our question is in m/s and velocity isn't speed is it but distance over time
    I suspect the problem is a typo. They aren't calling the unit for speed cm/s, they are calling the variable for speed "c".

    Also note:
    velocity = displacement over time
    speed = distance over time

    -Dan
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  10. #10
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    Quote Originally Posted by topsquark
    I suspect the problem is a typo. They aren't calling the unit for speed cm/s, they are calling the variable for speed "c".


    -Dan
    Dan therefore how would i do that question part ??

    Thanks for your input Nath
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  11. #11
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by nath_quam
    Dan therefore how would i do that question part ??

    Thanks for your input Nath
    I see nothing wrong with CaptainBlack's solution. It looks like the solution is x = 1 m and x = 3 m to me.

    -Dan
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  12. #12
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    Quote Originally Posted by topsquark
    I suspect the problem is a typo. They aren't calling the unit for speed cm/s, they are calling the variable for speed "c".

    -Dan
    If that is the intention it leaves v undefined.

    RonL
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  13. #13
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by CaptainBlack
    If that is the intention it leaves v undefined.

    RonL
    Well, that's why I suspect a typo. Note that v and c are next to each other on the keyboard. I suspect they meant to say "v m/s" instead of "c m/s."

    -Dan
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  14. #14
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    Quote Originally Posted by topsquark
    Well, that's why I suspect a typo. Note that v and c are next to each other on the keyboard. I suspect they meant to say "v m/s" instead of "c m/s."

    -Dan
    I had finaly realised that is what you mean in the car between work and
    home. I will add/modify the comment to the relevant post to clarify the matter.

    Thanks RonL
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  15. #15
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    Quote Originally Posted by CaptainBlack
    You are given:

    v=(2-x)^2,

    so for part (i):

    <br />
a=\frac{dv}{dt}=\frac{d}{dt}(2-x)^2=2(2-x)(-1)\frac{dx}{dt}=-2(2-x)^3<br />

    If as pointed out by nath_quam v is in cm/s but acceleration is wanted in
    m/s^2, in the above equation the acceleration is in cm/s^2 and
    needs to be divided by 100 to convert it into m/s^2. However as Topsquark
    points out there is probably a typo in the question and what appears as
    "c m/s", should read "v m/s" then the dimensions in the above are OK
    (unless someone else spots an error ).

    RonL
    i spotted an error we don't want to integrate answer should be 2x - 4
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