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!

Mechanics problem

  1. Jun 9, 2008 #1
    1. The problem statement, all variables and given/known data

    A cyclist of mass m experiences a force due to wind resistance equal to -kv^2 when travelling at speed v on a horizontal road. Show that her speed halves in a distance of 0.69m/k if she stops pedalling.

    2. Relevant equations


    3. The attempt at a solution
    ma = -kv^2
    i m not sure if i m right.
     
    Last edited: Jun 9, 2008
  2. jcsd
  3. Jun 9, 2008 #2


    well, you are right =)
    So you are able to do it now?
     
  4. Jun 9, 2008 #3
    lol :D but i don't get where to put 0.69m/k into the equation
    and does it mean a = 0 when she stops pedalling?
     
  5. Jun 9, 2008 #4

    tiny-tim

    User Avatar
    Science Advisor
    Homework Helper

    Hi suppy123! :smile:

    erm … you don't put 0.69m/k into the equation, you get it out of the equation! :rolleyes:

    And no, a = 0 only when v = 0.

    Hint: a = dv/dt. :smile:
     
  6. Jun 9, 2008 #5
    hi, tim ;)
    ma = -kv^2
    m(dv/dt) = -kv^2
    m(dv/dt) = -k(d/t)^2
    m(dv/dt) = -k d^2/t^2
    m(dv/dt) * 1/k^2 = -k * d^2/t^2 * 1/k^2
    m(dv/dt) / k^2 = -d/k * d/t^2
    i got it out ;)
     
  7. Jun 9, 2008 #6

    tiny-tim

    User Avatar
    Science Advisor
    Homework Helper

    oh suppy123 … a lot of that makes no sense at all! :redface:

    (d/t) doesn't mean anything.

    Nor does d^2/t^2 or d/k or d/t^2.
    Hint: m(dv/dt) = -kv^2,

    so mdv/v^2 = -kdt. :smile:
     
  8. Jun 9, 2008 #7
    oh, does m in 0.69m/k means mass over k?
    mdv/v^2 = -kdt
    m/k = -dt/dv/v^2
    m/k = -dt * v^2 /dv
     
  9. Jun 10, 2008 #8

    tiny-tim

    User Avatar
    Science Advisor
    Homework Helper

    integrate

    Hi suppy123! :smile:
    hmm … very good … but you're just juggling with fractions.

    you need to integrate.

    Hint: ∫mdv/v^2 = -∫kdt :smile:
     
  10. Jun 10, 2008 #9
    oo, but how do i find out the "t" from k
    or it is just -m/v = 0 ??
     
  11. Jun 10, 2008 #10

    tiny-tim

    User Avatar
    Science Advisor
    Homework Helper

    You find the "t" by integrating, until you get an equation in t.

    Fist step: What is ∫kdt?
     
  12. Jun 10, 2008 #11
    er,
    k= -ma/v^2
    k=-m(dv/dt)/v^2
    ∫kdt = -∫m(dv/dt)/v^2 dt = -∫m/v^2 dv = m/v
     
  13. Jun 10, 2008 #12

    tiny-tim

    User Avatar
    Science Advisor
    Homework Helper

    Yay! :biggrin:

    Now integrate the left-hand side … what is ∫kdt? :smile:
     
  14. Jun 10, 2008 #13
    ∫mdv/v^2 = -m/v = -∫kdt
    ∫kdt= m/v
     
  15. Jun 10, 2008 #14

    tiny-tim

    User Avatar
    Science Advisor
    Homework Helper

    Yes, I know!

    And I know you know!

    And you know I know you know! :rolleyes:

    But … now integrate the left-hand side! :smile:
     
  16. Jun 10, 2008 #15
    but i don't know the t
    the left-hand side which is....?
     
  17. Jun 10, 2008 #16

    tiny-tim

    User Avatar
    Science Advisor
    Homework Helper

    What who did next?

    D'oh!

    Hint:
    What is ∫kdt? :smile:
     
  18. Jun 10, 2008 #17
    0????????
     
  19. Jun 11, 2008 #18

    tiny-tim

    User Avatar
    Science Advisor
    Homework Helper

    … what kt did next …

    ∫kdt = kt + constant.

    So the original ∫kdt = -∫m dv/v^2 becomes:

    kt = m/v + constant. :smile:
     
  20. Jun 11, 2008 #19

    rl.bhat

    User Avatar
    Homework Helper

    Since you want the relation between distance and velocity, try this one.
    ma = -kv^2
    dv/dt = -k/m*v^2
    dv/ds*ds/dt = -k/m*v^2
    dv/ds*v =-k/m*v^2
    dv/ds = -k/m*v
    dv/v = -k/m*ds
    Now integrate. Put the initial condition to find the constant of integration.
     
  21. Jun 11, 2008 #20
    - m/k = -vt+vc
    m/k=vt-vc
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?



Similar Discussions: Mechanics problem
  1. A problem in mechanics (Replies: 12)

  2. Problem in mechanics (Replies: 12)

  3. Mechanics Problem (Replies: 2)

Loading...