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Homework Help: Magnitude of Velocity

  1. Aug 4, 2010 #1
    1. The problem statement, all variables and given/known data

    Given that:

    8y[tex]^{2}[/tex] + 4y = -4x + 6
    x + y = 4t

    where x and y are in metres, and t is in seconds.

    If y is positive, what is the magnitude of the velocity when t = 0.1875s?

    2. Relevant equations

    N/A

    3. The attempt at a solution

    I am a little unsure on how to start.

    I think I have to differentiate:

    8y[tex]^{2}[/tex] + 4y = -4x + 6

    Using implicit, then sub in x + y = 4t

    And solve for x and y.

    If all that is right, which I doubt it is. I then have to some how find the magnitude of velocity, which I am unsure how to do.

    Any and all help would be greatly appreciated. :D
     
    Last edited: Aug 4, 2010
  2. jcsd
  3. Aug 4, 2010 #2
    You're pretty close.

    Use implicit differentiation on the first equation, and solve for y' in terms of x and y.

    Also, solve for y in the first equation when x = 4(0.1875) - 7. Take the positive root of y since the problem says y > 0.

    Now plug in the positive root you found for y and x = 4(0.1875) - 7 into the equation you found for y' in the first step. This should be the velocity at t = 0.1875 s.
     
  4. Aug 4, 2010 #3
    what i basically see is that u have been given the equation of the trajectory of a particle and the relation between its x co-ordinate ad time is given.
    So what u do is:

    obtain, dx/dt=4, which is the velocity along x-axis,
    at t=0.1875s, u can find x from the 2nd equation and hence plug its value into the 1st equation to obtain y. Once u have the x and y values of the function at t=0.1875s,
    differentiate the 1st equation u wrote, with respect to 't',
    u will get dy/dt in terms of y and dx/dt. hence, dy/dt is the velocity along y axis.
    at t=0.1875s, plug the value of y u obtained, and u have already calculated dx/dt, so u can obtain dy/dt.

    So the magnitude of velocity of the particle at that instant is just {(dy/dt)^2+(dx/dt)^2]^0.5
     
  5. Aug 4, 2010 #4
    First of I must say, sorry that the equation was x + y = 4t. My bad I accidentally wrote x + 7 for some reason. T.T


    Ok so first I implicit differentiate:

    8y[tex]^{2}[/tex] + 4y = -4x + 6


    16y . [tex]\frac{dy}{dx}[/tex] + 4 . [tex]\frac{dy}{dx}[/tex] = -4


    [tex]\frac{dy}{dx}[/tex] = [tex]\frac{-4}{16y + 4}[/tex]

    Ok so now I solve for y in the second equation:

    x + y = 4t = [tex]\frac{3}{4}[/tex] = 0.75

    y = 0.75 - x


    Now I am kinda stuck again. Because I am unsure what to do. If you sub in that equation you will always have one variable in it. Not sure what to do.
     
    Last edited: Aug 4, 2010
  6. Aug 4, 2010 #5

    Dick

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    I would split this into two parts. First put your value for t in. Now you have two equations in two unknowns for x and y. Find them. Then differentiate d/dt. Now you have two equations in two unknowns for x'=dx/dt and y'=dy/dt. Use the value you found for y in the first part. Then the magnitude of the velocity is sqrt(x'^2+y'^2). I haven't tried to work it out, but does that make sense?
     
  7. Aug 4, 2010 #6
    x = 4t - y

    dx/dt = 4 (because then y would be a contant.)

    You want me to sub in x + y = 4t into the equation below????....????....

    8y² + 4y = -4x + 6

    8y² + 4y = -4(4t - y) + 6

    8y² + 4y = -16t + 4y + 6

    8y² = -16t + 6

    y = sqrt(-2t + 6/8)

    dy/dt = 0.5 x -2 / sqrt(-2t + 6/8) = - 1/sqrt(-2t + 6/8)

    I guess I will sub in t = 0.1875, and I get 1.63, square that and 4, and root it. I get 4.320 m/s. Which unfortunately is not right.
     
    Last edited: Aug 4, 2010
  8. Aug 4, 2010 #7

    Dick

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    These aren't partial derivatives. y isn't constant. It's a function of t. Differentiating x=4t-y gives you dx/dt=4-dy/dt.
     
  9. Aug 4, 2010 #8
    Ok so

    8y² + 4y = -4x + 6

    16y dy/dt + 4 dy/dt = -4 dx/dt

    (16y + 4) dy/dt = -4 . dx/dt

    sub in dx/dt = 4 - dy/dt

    (16y + 4) dy/dt = -4 (4 - dy/dt)

    (16y + 4) dy/dt = -16 + 4 dy/dt

    (16y) dy/dt = -16

    (y) dy/dt = -1

    dy/dt = -(1 / y)

    And

    dx/dt = 4 - dy/dt

    dx/dt = 4 + (1 / y)

    Now I we have dy/dt and dx/dt, I hope. :S

    I am unsure how to get the velocity???


    I know velocity = sqrt(x'^2+y'^2)

    but there is still the y in the equation, and I can't remove it using "x + y = 4t"

    because that just changes y into x, and vise versa, and you still get an x or a y in your answer.
     
    Last edited: Aug 4, 2010
  10. Aug 4, 2010 #9

    vela

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    Close, but you made a sign error right at the start when you differentiated -4x. Try fixing that first.
     
  11. Aug 4, 2010 #10
    fixed

    I am still a little stuck at the same point
     
  12. Aug 4, 2010 #11

    vela

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    If you can figure out what y equals at t=0.1875 s, you can calculate the two components. So following Dick's suggestion, set t=0.1875 s in the original equations, and you'll get

    [tex]8 y^2 + 4y = -4x + 6[/tex]

    [tex]x + y = 0.75[/tex]

    You have two equations and two unknowns. Solve for y.
     
  13. Aug 4, 2010 #12
    8y² + 4y = -4x + 6

    sub in : "x = 0.75 - y"

    8y² + 4y = -4(0.75 - y) + 6

    8y² + 4y = -3 + 4y + 6

    8y² = 3

    y = sqrt(3/8) - because y > 0

    If I sub that value into dx/dt and dy/dt

    dy/dt = -(1 / y)

    dy/dt = -(1 / sqrt(3/8)) = -1.633

    dx/dt = 4 + (1 / y)

    dx/dt = 4 + (1 / sqrt(3/8)) = 5.633

    velocity = sqrt( 5.633² + (-1.633)² ) = 5.865 m/s

    WHICH IS CORRECT!!! Thank you so so much :D
     
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