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Engineering kinematics

  1. Oct 4, 2015 #1
    i have found out the second answer of the PHYSICS problem but unable to solve the first part.i tried all possible ways.:-(
    a test projectile is fired horizontally in to a viscous liquid with a velocity Vo . The retarding force is proportional to the square of the velocity, so that the accelaration becomes , a=-kV^2.

    Derive an expression for the distance D travelled in the liquid and the corrosponding time required to reduce the velocity to Vo/2.Neglect any vertical motion.

    Ans. 1. D=0.693/K
    2. t=1/(kVo).

    (I solved for the second answer, but am unable to solve for the first one...)
    Last edited by a moderator: Oct 4, 2015
  2. jcsd
  3. Oct 4, 2015 #2


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    What have you tried?
  4. Oct 4, 2015 #3
    a= - kVo^2

    Since a=dv/dt

    Vo/t = - kVo^2
    or, t= - 1/(kVo)
    or, t = 1/(kVo) , as time can't be negative.

    Attached Files:

  5. Oct 4, 2015 #4


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    ... and, for d?
  6. Oct 4, 2015 #5
    This solution to the differential equation is incorrect. The differential equation reads:


    Note that there are V's on both sides of the equation. Do you know how to solve such an equation? If so, let's see.

  7. Oct 6, 2015 #6
    sorry then its totally wrong i dont know what to do now...plz advice ,,, the first part to find distance feels impossible to find , getting no clues where to start
  8. Oct 6, 2015 #7
    and tomorrow is my exam , god knows if this comes in exam , its in the suggestion which means it can come in exam , i am striving with this math for four days .......................... :-(
  9. Oct 6, 2015 #8
    Are you taking, or have you taken a course in differential equations? If so, then you know you can solve the differential equation in post #5 using "separation of variables."

  10. Oct 6, 2015 #9
    i have just started with calculas in class , and have not got in to depth yet so may be thats why not able to solve it , but how will i find distance using the differential equation
    . i am not taking any course but i am in std X
  11. Oct 6, 2015 #10
    I have to find the distance but how ? ?
  12. Oct 6, 2015 #11
    normal v^2 = u^2 - 2aS , is not working .
    where v= final velocity
    u= initial velocity
    S= distance

    what to do ? ? ? :-(
  13. Oct 6, 2015 #12
    dv/dt=- k v^2
    or, 1/v^2 dv = -k dt
    or, v^ -2 dv = - k dt
    integrating both sides we get
    v^(-2+1)/-2+1 = - kt
    or,v^-1 = kt
    or, t = 1/kv

    if v = Vo
    then , t = 1/kVo

    this is the second part i think but what about the 1st part . i am so much in dispair ...what to do plz help ?????
  14. Oct 6, 2015 #13


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    That equation only applies when acceleration is constant which isn't the case in this problem.

    See posts #5 and #8 and look up "separation of variables" as Chester suggests.
  15. Oct 6, 2015 #14
    This equation is incorrect. You forgot to include the constant of integration. You can determine the constant of integration by making use of the initial condition, v = Vo at t = 0.
    So, using this, what equation do you get for the velocity?

  16. Oct 8, 2015 #15
    Could you find the distance?
  17. Oct 8, 2015 #16
    You would have to next integrate the velocity to get the distance.
  18. Oct 9, 2015 #17
  19. Oct 9, 2015 #18
    Sorry I did mistake
  20. Oct 9, 2015 #19
    It should be as follows
  21. Oct 9, 2015 #20

    Attached Files:

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