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Instantaneous velocity and acceleration

  1. Oct 7, 2004 #1
    Hi,
    This is probably an easy question for some of you, but I do not Undrestand it..

    x=At^4 + (Bt + C)t^2 + D * sin(Et) where A, B, C, D, and E are constants. and I have to find instantaneous acceleration and velocity.

    I know Instantaneous velocity is
    v_x = dx/dt

    Instantaneous acceleration dv_x/dt

    But I do not know how to solve this equation..
     
    Last edited: Oct 7, 2004
  2. jcsd
  3. Oct 7, 2004 #2

    arildno

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    You're absolutely correct in which formulas you are to use!
    Now, let's find the velocity first:
    You've been given x(t), could you state precisely why you find differentiating x(t) looks hard? (Refer to specific terms you find troublesome)
     
  4. Oct 7, 2004 #3
    do I just fond the derivative,
    for example

    4At^3+3Bt^2+2Ct+Dcos(Et), something like this, cause Im not sure how it should look.
     
  5. Oct 7, 2004 #4

    arildno

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    A bit too fast, there!
    Let's take it slower:
    1) You know that if h(t)=f(t)+g(t), then we have:
    [tex]\frac{dh}{dt}=\frac{df}{dt}+\frac{dg}{dt}[/tex]
    that is, the derivative of a sum is the sum of the derivatives, right?
    Let's use that to split up our task:
    [tex]x(t)=At^{4}+Bt^{3}+Ct^{2}+D\sin(Et)[/tex]
    Hence, we have:
    [tex]\frac{dx}{dt}=\frac{dF}{dt}+\frac{dG}{dt}+\frac{dH}{dt}+\frac{dI}{dt}[/tex]
    where I've introduced:
    [tex]F(t)=At^{4},G(t)=Bt^{3},H(t)=Ct^{2},I(t)=D\sin(Et)[/tex]
    Now, as you can see, you have computed correctly the following terms:
    [tex]\frac{dF}{dt},\frac{dG}{dt},\frac{dH}{dt}[/tex]
    But what about [tex]\frac{dI}{dt}[/tex]?
     
  6. Oct 7, 2004 #5
    would it be sin(Et) only..
     
  7. Oct 7, 2004 #6

    arildno

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    Hint:
    Have you learnt the CHAIN RULE yet?
     
  8. Oct 7, 2004 #7
    yeah I did in Cal1, but havent touched it in a while
     
  9. Oct 7, 2004 #8
    Sin(Et)+Dcos(Et)? something like this, Ill go back and read my Calculus book.
     
  10. Oct 7, 2004 #9

    arildno

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    OK, that's the rule you need to use, in order to differentiate Dsin(Et) correctly.
    So, how would you proceed to do that?
     
  11. Oct 7, 2004 #10

    arildno

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    Your posts 8+5 have wrong suggestions.
     
  12. Oct 7, 2004 #11
    if I replace D, E with X, can I do that?

    then I would get

    xsin(xt)=sin(Xt)cos(Xt)?
     
  13. Oct 7, 2004 #12
    would the instantaneous velocity and acceleration be different in this question.
     
  14. Oct 7, 2004 #13

    arildno

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    NO!
    It seems to me you find the chain rule kind of hard to understand;
    that's fairly common, though. It IS rather difficult.
    Let's suppose you've got two functions, f(t) and g(t).
    You can now make yourself a new function, h(t), by substituting g(t) into all instances of "t" in f.
    That is, we write:
    h(t)=f(g(t))
    Now, the chain rule says:
    h'(t)=f'(g(t))g'(t).
    Let's be absolutely clear about what this MEANS, in terms of how to compute the derivative of h(t) (that is, h'(t)) correctly:
    1. Differentiate f(t) as usual, that is find [tex]\frac{df}{dt}[/tex]
    2. Substitute into all instances of "t" in your expression for [tex]\frac{df}{dt}[/tex] g(t)
    3. The expression you've now found, is f'(g(t)), that is, one of the factors in h'(t).
    4. Differentiate g(t) as usual, that is, find [tex]\frac{dg}{dt}[/tex] (or, which is the same, g'(t))
    5. Finally, multiply together f'(g(t)) (found in 2&3) and g'(t).
    You have now found h'(t)=f'(g(t))g'(t)

    To your problem:
    In order to use the chain rule, you must first find good choices for f(t) and g(t)!
    We have: [tex]I(t)=D\sin(Et)[/tex]
    A good choice of f(t) is:
    [tex]f(t)=D\sin(t)[/tex]
    (The reason why this is a GOOD choice, is that we know how to differentiate sine!)
    We therefore set:
    g(t)=Et
    We do this, because we now see that we may write:
    I(t)=f(g(t))
    Agreed?
    Hence, let's calculate [tex]\frac{dI}{dt}[/tex] according to the scheme above!
    1. [tex]\frac{df}{dt}=D\cos(t)[/tex]
    Agreed?
    2&3. [tex]f'(g(t))=D\cos(Et)[/tex]
    4. [tex]\frac{dg}{dt}=E[/tex]
    5. [tex]f'(g(t))g'(t)=DE\cos(Et)[/tex]
    Hence, we have:
    [tex]\frac{dI}{dt}=DE\cos(Et)[/tex]
     
    Last edited: Oct 7, 2004
  15. Oct 7, 2004 #14

    arildno

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    OOPS!
    Dreadfully sorry, wrote "cosine" at some places where I meant "sine".
    This is now edited.
     
  16. Oct 7, 2004 #15
    If is the answer for the instantaneous velocity, what is the difference between that and instantaneous acceleration.
     
  17. Oct 7, 2004 #16

    arildno

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    In order to find the acceleration, you must differentiate your new-found expression for the velocity.
     
  18. Oct 7, 2004 #17
    This is what I get,

    12At^2 + 6Bt + 2C - DEsin(Et)

    is this correct?
     
  19. Oct 7, 2004 #18

    arildno

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    Again, you're right about the 3 first terms; you haven't done the last term correctly, though (it's that damn chain rule again..:wink:)
     
  20. Oct 7, 2004 #19
    -DE^2sin(Et)?
     
  21. Oct 7, 2004 #20

    arildno

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    Precisely!
     
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