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!

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

Have something to add?