Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: Simple velocity vector problem; IS MY SOLUTION MANUAL WRONG OR AM I?

  1. Mar 30, 2008 #1
    Simple velocity vector problem; IS MY SOLUTION MANUAL WRONG OR AM I?!

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

    A particle starts from the origin with velocity 5i at t=0, and moves in the xy plane with a varying acceleration given by a= 6 Sqrt(t) j, where t is in s.
    a.) Determine the vector velocity of the particle as a function of time.

    2. Relevant equations
    Vi=5i; a = 6 sqrt(t)j
    Vf=Vi+at
    Vf = Vfx +Vfy = (Vix+axt)i+(Viy+ayt)j
    =(5+0)i + (0+6sqrt(t)*t)j
    =5i +6t^3/2 j

    The solution manual states the answer as Vf=5i +4t^3/2 j
    Whos right here?
     
  2. jcsd
  3. Mar 30, 2008 #2
    anyone? please? pretty please?
     
  4. Mar 30, 2008 #3

    tiny-tim

    User Avatar
    Science Advisor
    Homework Helper

    Hi HelpMeWIN123! :smile:

    No no no … that formula only works if a is constant! :frown:

    You must integrate … :smile:
     
  5. Mar 30, 2008 #4
    can you be more specific, integrate what, and integrate why?
     
  6. Mar 30, 2008 #5
    AHHH you're right, and a genius, and a scholar.
    Acceleration is not constant, so we can't use those formulas for constant acceleration.
    whenever they give you an a with respect to time, this is a variability acceleration, so in order to achieve V we must integrate a.

    Thanks.
    But now if i'm trying to find the position with respect to time;
    it would be : 5t i + [0 +0t + (1/2) (6 sqrt(t)*t^2) ]j
    isn't it?
    which gives me 3t^5/2 j? this isn't right either ahhhhhh since acceleration isn't constant..
    so i took the velocity vector 5i +4t^3/2j
    and differentiated it to achieve a, for each component giving me, a= 0i +6 sqrt(t)j which give me right back to where I was in terms of the previous plug-in. grr
     
    Last edited: Mar 30, 2008
  7. Mar 30, 2008 #6

    tiny-tim

    User Avatar
    Science Advisor
    Homework Helper

    ok … acceleration is rate of change of velocity.

    In other words: a = dv/dt.

    So if you integrate, you get ∫adt = ∫(dv/dt)dt = vf - vi.

    If a is constant, that's just the usual at = vf - vi formula.

    But in this case, a isn't constant.

    So … ? :smile:
     
  8. Mar 30, 2008 #7

    tiny-tim

    User Avatar
    Science Advisor
    Homework Helper

    ah … just seen your last post.

    ok … same thing … speed is rate of change of distance.

    In other words: v = dy/dt.

    So if you integrate, you get ∫vdt = ∫(dy/dt)dt = yf - yi.

    So just integrate v … :smile:

    (You don't like integrating, do you? :rolleyes:

    I think it would be best if you always wrote out the formulas, like v = dy/dt, for yourself at the start.)
     
  9. Mar 30, 2008 #8
    so i just integrated that further and got the answer., integrated the velocity rather.
     
  10. Mar 30, 2008 #9
    Yeah you're right, I guess here I just didn't understand how to differentiate (figuratively) the different of when I have constant a or constant v versus not, the thing is once I have the velocity that I got from integrating my acceleration, why can't i just plug that into my kinematics equation? kinematics eq. says nothing about having constant v
     
  11. Mar 30, 2008 #10

    tiny-tim

    User Avatar
    Science Advisor
    Homework Helper

    Hi HelpMeWIN123! :smile:

    If in doubt … integrate anyway … even if it's a constant, you'll still get the correct result! :smile:
    Sorry … what kinematics equation? :confused:
     
  12. Mar 30, 2008 #11
    the kinematics equation i used upstairs; yf = yi +viy *t +1/2 ay t^2
    also how does ∫(dy/dt)dt = yf - yi
    could i just shortcut and say ∫(dy/dt)dt = yf?
    I think i'm utterly lost here...isn't dy/dt already vf-vi when change in time approaches 0
     
    Last edited: Mar 30, 2008
  13. Mar 30, 2008 #12

    tiny-tim

    User Avatar
    Science Advisor
    Homework Helper

    This kinematics equation only works for constant a, anyway. :frown:
    You can only "shortcut" if yi (the initial distance) is zero.

    Do you understand the equation ∫(dy/dt)dt = yf - yi ?
     
  14. Mar 30, 2008 #13
    well i was tryign to make sense of it in the fact that
    V= dy/dt
    Vdt = dy, and dy = yf-yi or change in y
    then ∫V dt = yf - yi is where i get stuck because how you can just integrate one side and how do you have the wherewithall to do this.
     
  15. Mar 30, 2008 #14

    tiny-tim

    User Avatar
    Science Advisor
    Homework Helper

    No … it's because y = ∫(dy/dt)dt.

    (Perhaps that looks more familiar as ∫y´= y ?)

    So ∫(dy/dt)dt between a and b

    = [y] between a and b,

    = y(a) - y(b) …
    which in dynamics we usually write yf - yi. :smile:
     
  16. Mar 30, 2008 #15
    ahhh thank you sOOOO MUCH!!!!!
     
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook