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Kinematics: finding the average acceleration

  1. Jan 24, 2013 #1
    1. The problem statement, all variables and given/known data
    The velocity of a particle moving along the x axis varies in time according to the expression

    v(t)=α+βt2

    α=46.4m/s
    β=4.25m/s3
    t is in seconds

    Find the average acceleration in the time interval from t=0 to 3.63s.
    Answer in units of m/s2




    2. Relevant equations
    I really do not have the slightest idea as to what equation to use for this. I know I need to find the average acceleration which is a=Δv/Δt, however, I don't recognize how to get the initial and final velocities out of that problem. It's already given me the initial time and final time so obviously 3.63s-0s=3.63s but as previously stated, how do I pull the initial and finals out of that? (Even if that is the way to go about it.) I have tried a few things and my results of those will be under this.




    3. The attempt at a solution
    First I got 28.20986915 just by plugging and chugging. I'm still working on it but I am in desperate need of a quick lesson on how to solve this problem and the concepts of it.
    Please, no answers; I only want to know and understand how to solve this.
    Thank you.

    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Jan 24, 2013 #2

    cepheid

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    Hello hrodger,

    Do you know what a function is? The velocity is a function of time. This means that you can plug in different values of t as the argument to the function, and different values of v result. What is the initial speed v(0)? What is the final speed v(3.63)? So, what is the difference between them?
     
  4. Jan 24, 2013 #3
    Thanks for the reply! To answer your question about functions, no, not particularly; I've worked with them a little bit in college algebra last semester, but that's it. And my final velocity (3.63) - my initial velocity (0) = 3.63s, I thought that the 3.63s was my time, not velocity? I must have misunderstood it.
     
  5. Jan 24, 2013 #4

    tms

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    You can't get 10 significant figures when you start with only 3.
     
  6. Jan 24, 2013 #5

    cepheid

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    The simplest way to think about it (without even worrying about the term "function") is that 't' is the "independent" variable. It doesn't depend on anything else. It varies independently.

    v is the "dependent" variable, because its value "depends" on the value of t.

    For every value of t (time) there is a corresponding value of v (velocity). That's what a function is, it's a mapping from one set of values to another set. As an analogy, you can think of a function being like a machine that takes an "input" (t) and according to some algorithm (procedure), spits out some output (v). In this case, the mapping is given by the following mathematical expression:

    v(t)=α+βt2

    So, for every value of t, you can find the corresponding value of v just by evaluating this expression. The function notation with the parentheses is meant to indicate this relationship. The thing inside the parentheses is called the "argument" to the function. It's the independent variable. Whatever you put in the parentheses is what you are passing as an input to the "machine." So, if you pass 0 as the value: v(0), then you are computing the value of v when t = 0, and you would do that just by setting t = 0 in the mathematical expression for the function:

    v(0)=α+β(0)2 = α

    If you pass 3.63 as the argument to the function: v(3.63), then you are computing the value of v when t = 3.63.
     
  7. Jan 24, 2013 #6
    Wasn't my final answer. I needed to get either the final velocity or initial velocity out of that equation. What my issue was is that I didn't know which. I asked a friend and he helped me figure it out. My final answer was 15.4275000 or 15.4m/s2 as the sig fig would require.
     
  8. Jan 24, 2013 #7
    I cannot begin to express my gratitude for the time you spent in explaining this to me. That is honestly the most clear-cut and simple way I have heard functions described. Thank you very much. The problem has been solved and I shouldn't ever have issues with functions again thanks to you, my friend. Good night!
    "If you can't explain it simply, you don't understand it well enough."
    Albert Einstein
     
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