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Homework Help: Need help with a Very Hard Kinematics Qn

  1. Jan 13, 2009 #1

    physicsnoob93

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    1. The problem statement, all variables and given/known data

    Imagine a tank traveling at constant speed v1 along direction AB and a missile is chasing the tank with constant speed v2. The direction of motion of the missile is always pointing toward the tank. At the t the tank is at position F and the missile is at position D, where FD is perpendicular to AB and FD = L. What is the acceleration of the missile at this instant?

    A---------------F---------------B
    ************|
    ************|
    ************|
    ************|
    ************D

    I have no clue, i have spent hours on this qn. I tried drawing the initial and final velocity vectors for the missile but i have no idea how to continue from there.

    Thanks if anyone can help.

    2. Relevant equations
    ?


    3. The attempt at a solution
     
    Last edited: Jan 13, 2009
    physicsnoob93, Jan 13, 2009
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  3. Jan 13, 2009 #2

    chrisk

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    Acceleration is a change in velocity with respect to time. The missle velocity relative to the tank is v2 - v1. There is no linear accleration in this problem. Are you trying to solve for the angular acceleration?
     
    chrisk, Jan 13, 2009
  4. Jan 13, 2009 #3

    Thaakisfox

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    Its best to use relative coordinates.
    Lets fix our system to the tank, and use polar coordinates, so r will be the instantaneous distance between the missile and the tank, and \phi will be the angle between this and the initial position.
    Our equations are then:

    [tex]\dot r=v_2-v_1\sin\varphi[/tex]

    and

    [tex]r\dot\varphi=v_1\cos\varphi[/tex]

    Now we know that the acceleration in polar coordinates is:

    [tex]\vec a = (\ddot r -r\dot\varphi^2)\vec e_{r} + (2\dot r\dot\varphi + r\ddot\varphi)\vec e_{\varphi}[/tex]

    Using the equations, calculate the acceleration, and after you are done put [tex]\varphi=0[/tex] and r=L, and you will get it at the beginning.

    It turns out to be:

    [tex]a=\frac{v_1v_2}{L}[/tex]

    (I be wrong, because I just scribbled it up quickly... :D)
     
    Thaakisfox, Jan 13, 2009
  5. Jan 14, 2009 #4

    physicsnoob93

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    Thanks for the replies. I solved it using similar triangles and i got v1v2/L too.

    My Solution is:

    If theres a triangle drawn for the distance,

    Theres the adjacent with l and the opposite to be V1*dt

    If theres a triangle for velocity,
    The adjacent is V2, the opposite is a*dt
    Then we get

    V1dt/l = adt/V2
    Simplifying,
    a = V1V2/l

    Thanks though :D
     
    physicsnoob93, Jan 14, 2009
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