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

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Homework Statement


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)

Homework 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

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.
 
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Answers and Replies

  • #2
cepheid
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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
That seems a bit silly, because you would have

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

on the right hand side.
 
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  • #3
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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?!
 
  • #4
Cyosis
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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.
 
  • #5
cepheid
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So how do you solve it in a way that's not silly?!
Just what I said. Acceleration is the second derivative of the position function, which you have been given already.
 
  • #6
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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?
 
  • #7
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It worked when I put -a! :D

So this is solved now, but I can't edit the title
 

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