1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Need help with a Very Hard Kinematics Qn

  1. Jan 13, 2009 #1
    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?


    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
  2. jcsd
  3. Jan 13, 2009 #2
    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?
  4. Jan 13, 2009 #3
    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]



    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:


    (I be wrong, because I just scribbled it up quickly... :D)
  5. Jan 14, 2009 #4
    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
    a = V1V2/l

    Thanks though :D
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?

Similar Discussions: Need help with a Very Hard Kinematics Qn
  1. Homework help very (Replies: 32)

  2. Help, Kinematics (Replies: 2)