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Using a graph to find velocity and position at a said time

  1. Jul 21, 2011 #1
    1. The problem statement, all variables and given/known data

    Consider the plot below describing the acceleration of a particle along a straight line with an initial position of −15 m and an initial velocity of −3 m/s.

    [​IMG]

    A) What is the velocity at 5 s? Answer in units of m/s.

    B) What is the position at 5 s? Answer in units of m.

    2. Relevant equations

    v = v0 + (a)(t)

    3. The attempt at a solution

    A) Because the acceleration is obviously not constant, I added the periods of constant acceleration, from 0 to 2 seconds and 2 to 5 seconds as so:

    v= -3 + (7)(2) = 11
    v= 11 + (1)(5) = 16
    ---
    27 (answer)

    I'm pretty sure that I'm wrong, and I'm confused about a couple of things:
    1. Is the initial velocity always going to be -3 m/s, or is it going to be the velocity that was the end point of the other equation? Because above, I got 11 m/s for the first equation - does that mean that 11 m/s will be my initial velocity for the second equation (when I'm doing the second part of the graph) as well? Or will it be -3 m/s?

    2. I know this is a stupid question, but is the time variable a change in time, in other words, the final time minus the time you're starting at? Like looking at the graph, from the interval from 2 to 5 seconds when the acceleration is constant, is "t" 3 seconds, the change in time, or is it 5 seconds, the final time the acceleration is constant? I'm confused as to what to use.

    Once I get that squared away, I'll be able to focus on part "B" of the problem.

    Thanks for the help!
     
  2. jcsd
  3. Jul 21, 2011 #2

    gneill

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    Staff: Mentor

    Yes, 11 m/s will be the initial velocity for the second stage.
    For the motion equation v(t) = vo + a*t you always assume that time begins at zero. So for your calculations you want to use the Δt. That is, you "reset the clock" to zero at the beginning of each stage and then the variable t becomes the elapsed time during that stage.
     
  4. Jul 21, 2011 #3
    So I'd change my second equation accordingly, and this would be the result, right?

    v= -3 + (7)(2) = 11
    v= 11 + (1)(3) = 14

    Total velocity = 25 m/s
     
  5. Jul 21, 2011 #4

    PhanthomJay

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    yes
    no
    why are you adding the velocities?
    no
    yes
    which one do you think?
    t is the change in time, good question.
     
  6. Jul 21, 2011 #5
    I'm adding the velocities because it isn't constant over the entire time, but it is constant at different intervals, so to find the velocity at a given time, I would add all the constant velocities up to that given point, right? If not, what would I do?
     
  7. Jul 21, 2011 #6

    gneill

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    Staff: Mentor

    No, don't add the final velocities from each stage. The final velocity at the end of the second stage *is* the final overall velocity.
     
  8. Jul 23, 2011 #7
    Thanks for the help...now for finding the position at 5s, the second part of the problem I listed above, I used this equation:

    x-x0 = v0t + .5(a)(t)^2

    For 0 to 2 seconds my equation was: x-(-15) = -3(2) + .5(7)(2)^2 = -7
    Then for 2 to 5 seconds, this was my equation: x-(-7) = 14(3) + .5(1)(3)^2 = 39.5

    However, 39.5 is coming out as the wrong answer...can anyone help me with what I'm doing wrong?
     
  9. Jul 23, 2011 #8

    gneill

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    Staff: Mentor

    For the second stage the "initial" velocity should be the velocity at the start of the stage, not the end of the stage. You've used 14 m/s as that velocity, which is the end-of-stage velocity.
     
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