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Math Help - kinematics/calculus

  1. #1
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    kinematics/calculus

    hi, here is a paraphased question i am having trouble with:

    A vehicle's mass (including passenger) = 400 kg
    after travelling 400 m the car stops accelerating. At this instant, the car's brakes are applied, and in addition, a small parachute opens at the rear to slow the car down.
    The retarding force applied by the brakes including friction is 5000 N. The retarding force due to the parachute is 0.5v^2 N where v m/s is the velocity of the car x meters beyond the 400 m mark.
    a) if a m/s^2 is the acceleration of the car during the retardation stage, write the equation of motion for the dragster during this stage:
    now what i sorta got was this:
    .5v^2 + 5000 = 400a
    or (1/2)v^2 = 400a - 5000
    is this right?

    b) By choosing an appropriate derivative form for acceleration, show that the differential equation relating v to x is:
    dv/dx = -(10^4 + v^2)/(800v)

    Now, im thinking that there is a form for acceleration which goes:
    d((1/2)v^2)/dx = a
    which looks awfully similar to my (1/2v^2)
    but im not too sure where to take it to show that equation.

    Any help?
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  2. #2
    Grand Panjandrum
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    Quote Originally Posted by scorpion007 View Post
    hi, here is a paraphased question i am having trouble with:

    A vehicle's mass (including passenger) = 400 kg
    after travelling 400 m the car stops accelerating. At this instant, the car's brakes are applied, and in addition, a small parachute opens at the rear to slow the car down.
    The retarding force applied by the brakes including friction is 5000 N. The retarding force due to the parachute is 0.5v^2 N where v m/s is the velocity of the car x meters beyond the 400 m mark.
    a) if a m/s^2 is the acceleration of the car during the retardation stage, write the equation of motion for the dragster during this stage:
    now what i sorta got was this:
    .5v^2 + 5000 = 400a
    or (1/2)v^2 = 400a - 5000
    is this right?
    This looks OK (accept the acceleration is in the opposite direction to the
    velocity and in the direction of decreasing x so we ned a - sign here), but
    given what the next part of this question is I would arrange it differently,
    like:

    a = -(v^2 + 10000)/800.

    b) By choosing an appropriate derivative form for acceleration, show that the differential equation relating v to x is:
    dv/dx = -(10^4 + v^2)/(800v)

    Now, im thinking that there is a form for acceleration which goes:
    d((1/2)v^2)/dx = a
    which looks awfully similar to my (1/2v^2)
    but im not too sure where to take it to show that equation.

    Any help?
    As I had above:

    a = -(v^2 + 10000)/800,

    or:

    dv/dt = -(v^2 + 10000)/800

    but dv/dt = dv/dx dx/dt = v dv/dx, so

    v dv/dx = -(v^2 + 10000)/800,

    rearranging:

    dv/dx = -(v^2 + 10000)/(800v)

    RonL
    Last edited by CaptainBlack; October 15th 2006 at 09:06 AM.
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  3. #3
    Forum Admin topsquark's Avatar
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    Ummm....

    CaptainBlack is correct, I'm just not sure where he's going with it.

    I would use

    a = v*(dv/dx) = [(1/2)d(v^2)/dx]

    as you were originally thinking.

    Then:

    a = -(v^2 + 10000)/800

    v(dv/dx) = -(v^2 + 10000)/800

    dv/dx = -(v^2 + 10000)/(800v)

    -Dan
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  4. #4
    Grand Panjandrum
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    Quote Originally Posted by topsquark View Post
    Ummm....

    CaptainBlack is correct, I'm just not sure where he's going with it.
    What I am trying to do is start from where scorpion007 is and reach
    what he is asked to derive (if you ignore that I had typed da/dx
    throughout where I meant dv/dt )

    RonL
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  5. #5
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by CaptainBlack View Post
    What I am trying to do is start from where scorpion007 is and reach
    what he is asked to derive (if you ignore that I had typed da/dx
    throughout where I meant dv/dt )

    RonL
    That was what I was having the problem understanding.

    -Dan
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  6. #6
    Grand Panjandrum
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    Quote Originally Posted by topsquark View Post
    That was what I was having the problem understanding.

    -Dan
    What is particulary annoying about that is that on the paper next to the
    computer where I outlined what I am doing it does read dv/dt

    RonL
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  7. #7
    Senior Member
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    thanks guys, i get it now
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