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!

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

    Bystander

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member
    2016 Award

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

    Since a=dv/dt

    So,
    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

    Bystander

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member
    2016 Award

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

    $$\frac{dV}{dt}=-kV^2$$

    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.

    Chet
     
  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."

    Chet
     
  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
    a=accelaration
    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

    CWatters

    User Avatar
    Science Advisor
    Homework Helper

    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?

    Chet
     
  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
    IMG_20151009_155800.jpg
     

    Attached Files:

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

Have something to add?
Draft saved Draft deleted



Similar Discussions: Engineering kinematics
  1. Kinematics Question (Replies: 6)

Loading...