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!

Dymanics, F=ma, F function of position & velocity (it's really important)

  1. May 25, 2009 #1
    1. The problem statement, all variables and given/known data
    The proposed design for an energy- absorbing bumper for a car exerts a decelerating force of magnitude bs + cv on the car when it collides with a rigid obstacle, where s is the distance the car travels from the point where it contacts the obstacle and v is the car's velocity. Thus the force exerted on the car by the bumper is a function of the car's position and velocity.

    (a) Suppose that at t = 0 the car contacts the obstacle with initial velocity u. Prove that the car's position is given as a function of time by s (equation provided below). To do this, first show that this equation satisfies Newton's second law. Then confirm that it satisfies the initial conditions s = 0 and v = u at t = 0

    (there's also (b) but maybe I could do it by myself if this is solved)

    2. Relevant equations

    F = bs + cv

    [tex] s = \frac{u}{2h}[e^{-(d-h)t} - e^{-(d+h)t}] [/tex]

    [tex] d = \frac{c}{2m} [/tex]

    [tex] h = \sqrt{d^2 - \frac{b}{m}} [/tex]

    b and c are constants

    3. The attempt at a solution
    Well I first tried substituting with s in bs + cv to try and simplify it to ma then I realized there wouldn't be an a in the left hand side of the equation, so I substituted a with dv/dt and I integrated both sides, but I was still unable to simplify it (because of the es!) - (Edit: I obviously couldn't have integrated correctly because I can't separate the variables). Is this even the way to go about it? I tried starting with bs + cv = ma and again substituting with a = v dv/ds but I was unable to separate the variables. Please help? It's important.
    I also tried differentiating s with respect to time and then substituting v in bs + cv with the result, but I still can't simplify.
     
    Last edited: May 25, 2009
  2. jcsd
  3. May 25, 2009 #2

    cepheid

    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    That seems a bit silly, because you would have

    [tex] a = \frac{d^2s}{dt^2} [/tex]

    on the right hand side.
     
    Last edited: May 25, 2009
  4. May 25, 2009 #3
    I was trying with a = v dv/ds, couldn't separate the variables.
    Edit: No sorry, first time I tried a = dv/dt, and I'd already substituted s with f(t) so there was no s in the right hand side. I probably did something wrong, because I still can't separate the variables.

    So how do you solve it in a way that's not silly?!
     
  5. May 25, 2009 #4

    Cyosis

    User Avatar
    Homework Helper

    The equation of motion you need to solve is:

    [tex]
    m\frac{d^2s}{dt^2}=bs+cv=bs+c \frac{ds}{dt} \Rightarrow m\frac{d^2s}{dt^2}-c \frac{ds}{dt}-bs=0[/tex]

    Does this differential equation look more familiar?

    Edit: Reading your question again you do not have to solve this yourself. All you have to do is check if the given solution is a solution to the differential equation given above given the boundary values in your problem.
     
  6. May 25, 2009 #5

    cepheid

    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    Just what I said. Acceleration is the second derivative of the position function, which you have been given already.
     
  7. May 26, 2009 #6
    Thanks a lot, I get it now... But I still can't do it :(
    Just to make sure this is what I'm supposed to be doing, I get v and a then I simplify this

    [tex]
    m\frac{d^2s}{dt^2}-c \frac{ds}{dt}-bs
    [/tex]

    And then the co-efficients of each e are supposed to add up to zero, right? They don't, I spent hours yesterday doing it over and over by hand, and today on CAS and it's not zero. The co-efficient of

    [tex] e^{ht-dt}[/tex]

    is

    [tex] \frac{bu}{2\sqrt{\frac{c^2}{4m^2}-\frac{b}{m}}} + cu - \frac{\sqrt{\frac{c^2}{4m^2}-\frac{b}{m}}mu}{2} - \frac{3uc^2}{8\sqrt{\frac{c^2}{4m^2}-\frac{b}{m}}m} [/tex]

    I realize this is kinda trivial but somehow my grade in introductory mechanics is hanging on it. What am I missing?
     
  8. May 27, 2009 #7
    It worked when I put -a! :D

    So this is solved now, but I can't edit the title
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Dymanics, F=ma, F function of position & velocity (it's really important)
  1. Deriving F=ma? a vs F? (Replies: 6)

  2. F=ma zipline problem (Replies: 7)

Loading...