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Vectors of Newton's equation

  1. Jan 15, 2016 #1
    1. The problem statement, all variables and given/known data
    So my textbook states that F = ma = m(d2r/dt2) can be integrated to give r = r0 + tv0 when F = 0.

    2. Relevant equations


    3. The attempt at a solution

    I've tried rewriting it as F = m(dv/dt) and integrating that to give Ft = mv + c but this is obviously not in the right form. I've also tried using F = dP/dt but that gives effectively the same answer as above. I have no idea how to get the solution given!
     
  2. jcsd
  3. Jan 15, 2016 #2

    Student100

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    When there is no force, naturally there is no acceleration. The result your textbook shows is just the basic kinematic equation without the acceleration term.

    Can you show the steps for your work? How're you actually setting things up? Are you using your initial condition?

    So for the work shown, what is Ft and mv? What happens when you take the integral of the constant? What is that constant?
     
    Last edited: Jan 15, 2016
  4. Jan 15, 2016 #3

    Ray Vickson

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    The vector ##\vec{r}(t) = (x(t),y(t),z(t))##, where x, y and z are its three components. What kind of functions x(t), y(t) and z(t) have second derivatives that are identically 0?
     
  5. Jan 16, 2016 #4
    All function such as 2t have a second derivative of 0 but i don't know where to go with that?
     
  6. Jan 16, 2016 #5
    F dt = m dv
    Ft = mv + c
    When F = 0 , c = -mv
    Ft = mv - mv = 0 <--- This obviously isn't right but i don't know what else to do!
     
  7. Jan 16, 2016 #6

    Samy_A

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    Why is Ft=0 not right when F=0?
     
  8. Jan 16, 2016 #7
    Well the equation makes sense when F = 0. But i still can't manipulate it into the form stated in the book.
     
  9. Jan 16, 2016 #8

    Samy_A

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    You have ##c+mv=0##, or ##v=-c/m##.
    What does this tell you about ##v##? (By the way, Don't forget that ##v## is a vector, as Ray mentioned. So better write ##\vec v##).

    What can you say about ##\frac{d \vec r}{dt}##?
     
  10. Jan 16, 2016 #9
    so this shows that t##\vec v## is a constant so ##\frac{d \vec r}{dt}## so would integrate to give -c/mt = v0t?
     
  11. Jan 16, 2016 #10

    Samy_A

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    t##\vec v## is not constant, ##\vec v## is constant.
    Integrating ##\frac{d \vec r}{dt}## should give you ##\vec r## as a function of ##t##.
     
  12. Jan 16, 2016 #11

    Ray Vickson

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    Are you telling us that you don't know where to go with that, or are you asking us if you don't know where to go with that (which is the effect of using a "?" at the end)?

    Anyway, the issue is: if ##d^2 f(t)/dt^2 = 0## for all ##t##, what is the most general possible form for ##f(t)##? Well, if you set ##g(t) = df(t)/dt##, you have ##dg(t)/dt = 0## for all ##t##. What is the most general form of a function whose derivative is identically equal to 0? Once you have answered that, you have a formula for ##df/dt##, which you can then integrate to get ##f(t)##.
     
    Last edited: Jan 16, 2016
  13. Jan 16, 2016 #12

    Student100

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    You aren't using your initial condition, and it seems like you don't know how to actually take the integral. Is the vector thing confusing you?

    ##\vec{F} = m\vec{a}## is at it's heart a differential equation. Lets just look at the x direction for now: ##F_x = ma_x## Now lets use the initial condition of no net force: ##0 = ma_x## We know ##a_x = \frac{d^2x(t)}{dt^2}## so go ahead and make that substitution: ##0 = m\frac{d^2x(t)}{dt^2}## Now integrate this looking at the definite integral from 0 to t so we can nicely define C. Whats going to happen to the mass, what expression will you end up with? This should help you answer Ray's question too I hope.
     
  14. Jan 16, 2016 #13
    I think I'm getting confused with the integration of a second order differential as I haven't done it in a while. I get t2/2 = mx +c but I'm really not sure if this is correct!
     
  15. Jan 16, 2016 #14

    Student100

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    Not correct at all.

    Show me what you're doing from ##0 = m\frac{d^2x(t)}{dt^2} ##
     
  16. Jan 17, 2016 #15
    0 = mdv(t)/dt
    ∫dt = m∫dv
    t = mv
    t = mdx/dt
    ∫txt = m∫dx
    t2/2 = mx + c
     
  17. Jan 17, 2016 #16

    Student100

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    First how does 0dt become just dt. Next, you just seem to be dropping the second derivative. You should also look at the definite integral.

    $$\int_0^t0dt = m\int_0^t\frac{d^2x(t)}{dt^2}dt$$

    Do you know how to take this? What's the fundamental theorem of calculus say?

    Further, for the mass we assume two things: It's both non-zero, and doesn't change with time. That's why we can pull it out of the integral. It also how we can get rid of the term all together.
     
  18. Jan 17, 2016 #17

    Ray Vickson

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    You are seriously over-thinking the problem. You do not need to know much about differential equations and the like; you just need to remember that a derivative of a function g(t) at t is the slope of the tangent line to the graph of y = g(t). So, if dg(t)/dt = 0 for all t, what does the graph of y = g(t) look like? What does that tell you about the nature of the function g(t)?
     
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