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

Variable acceleration question

  1. Jul 29, 2009 #1
    hi all,

    i think this should be simple but i just cant figuring it out

    i basically want to move a particle at rest from point A to point B by applying a force, F(t).
    the particle should be at point B when t = T

    knowing at that F = m * a(t), i m assuming m = 1, which means F(t) = a(t)

    i dont want the particle to swing around point B, meaning i want it to settle to rest at point B very quickly, this means velocity at point B should be 0, v(T) = 0

    so assuming d=distance, v=velocity, a=acceleration, i created the following condition below for my question:

    d(0) = 0
    v(0) = V0
    a(0) = A0

    d(T) = B
    v(T) = 0
    a(T) = 0

    i m thinking that many a(t) will satisfy the above boundary conditions, like a linear a(t) in the form of
    a(t) = c * t + A0

    then i got stuck......

    any help would be great......
    thanks in advance
     
  2. jcsd
  3. Jul 29, 2009 #2

    rcgldr

    User Avatar
    Homework Helper

    If a(t) = c * t + A0, then what are v(t) and d(t)?
     
    Last edited: Jul 30, 2009
  4. Jul 29, 2009 #3

    diazona

    User Avatar
    Homework Helper

    You're right on track so far, especially with the observation that there will be many solutions. You're going to need some sort of extra condition to determine the acceleration uniquely. For example, you could assume that it's linear, or sinusoidal, or you could try to find the pattern of acceleration that will produce the minimum amount of work. Or whatever. But once you figure out what kind of acceleration you want, it's really just a math problem.

    P.S. You might have to do some fiddling to satisfy all 6 of your boundary conditions.
     
  5. Jul 30, 2009 #4

    rcgldr

    User Avatar
    Homework Helper

    Note for the given problem statement, A0, V0, and B are related to each other in such way that only two of them can be variable, and the third value will be determined from the other two. For example, if you want A0 and V0 to be the inputs, then B is a calculated value.
     
  6. Jul 30, 2009 #5
    here is my approach

    i got this from some physics book
    1) D(t) = 1/2 * a(t) * t2 + v0 * t + d0

    differentiating 1) with respect to t, i get
    2) v(t) = dD/dt = 1/2 * t * a' + t * a + v0

    differentiating 2) with respect to t, i get
    3) a = a(t) = dv/dt = 2 * t * a' + 1/2 * t2 * a'' + a

    from 3) i get a 2nd order DE
    4) a'' = -4/t * a

    solving 4) i get
    a(t) = c1/t3 + c2

    which means that acceleration is infinity at a(0), which violates my boundary condition of
    a(0) = A0

    am i in the right direction?
    thanks in advance
     
  7. Jul 30, 2009 #6

    rcgldr

    User Avatar
    Homework Helper

    That is the equation for constant acceleration, a(t) = A0.

    You've already stated the equation for acceleration,

    a(t) = c t + A0

    Start with that and integrate to get v(t), then integrate again to get d(t). For the constants of integration, you've already defined them as V0 for v(0) and 0 for d(0).
     
  8. Aug 1, 2009 #7

    rcgldr

    User Avatar
    Homework Helper

    For anyone that was curious:
    Note that A0 is negative, B is a determined value.

    A(t) = t c + A0
    V(t) = 1/2 t2 c + t A0 + V0
    D(t) = 1/6 t3 c + 1/2 t2A0 + t V0 + 0

    For some t != 0, A(t) = 0, V(t) = 0, and D(t) = B (to be determined)

    cancel c term to solve for T:
    1/2 T A(T) = 1/2 T2 c + 1/2 T A0
    V(T) - 1/2 t A(T) = 1/2 T A0 + V0 = 0
    T = -2 V0 / A0

    use T to solve for c:
    A(T) = T c + A0 = (-2 V0/A0) c + A0 = 0
    c = A02 / (2 V0)

    use T and c to solve for B
    D(t) = 1/6 T3 c + 1/2 T2A0 + T V0 + 0 = B
    1/6 (-2 V0/A0)3 (A02 / (2 V0)) + 1/2 (-2 V0/A0)2 A0 + (-2 V0/A0) V0 = B
    -2/3 (V02 / A0) + 2 (V02 / A0) -2 (V02 / A0) = B
    B = -2/3 (V02 / A0)

    alternate method for B (cancel c term, then A0 term, to solve for B versus t):
    cancel c term:
    V(t) = 1/2 t2 c + t A0 + V0
    D(t) = 1/6 t3 c + 1/2 t2A0 + t V0 - B = 0
    1/3 t V(t) = 1/6 t3 c + 1/3 t2 A0 + 1/3 t V0 = 0
    D(t) - 1/3 t V(t) = 1/6 t2 A0 + 2/3 t V0 - B = 0
    cancel A0 term:
    V(t) - 1/2 t A(t) = 1/2 t A0 + V0 = 0
    1/3 t (V(t) - 1/2 t A(t)) = 1/6 t2 A0 + 1/3 t V0 = 0
    (D(t) - 1/3 t V(t)) - (1/3 t (V(t) - 1/2 t A(t))) = 1/3 t V0 - B = 0
    B = 1/3 t V0
    solve for t = T
    B = 1/3 T V0 = 1/3 (-2 V0 / A0) V0 = -2/3 (V02 / A0)
     
    Last edited: Aug 2, 2009
  9. Aug 2, 2009 #8
    I'm pretty sure a constant acceleration is still within the family of initial solutions.
     
  10. Aug 2, 2009 #9

    rcgldr

    User Avatar
    Homework Helper

    Only if A0, V0, and B are zero, the object never moves, and t doesn't matter. Otherwise if A0 0 at t = 0, then later A0 = 0 at t = T, then acceleration has to change. Remove the requirement that A0 = 0 at t = T, then constant acceleration works.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook