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Finding the magnitude of a particle from a graph?

  1. Jul 24, 2011 #1
    1. The problem statement, all variables and given/known data

    Consider the acceleration of a particle along a straight line with an initial position of 0 m and an initial velocity of 0 m/s.

    5b9pgo.png

    1) Calculate the magnitude of the displacement after the car travels the first 5 s.
    Answer in units of m.

    2) How far does the car travel from 9 s to 13 s, if it does not experience any acceleration during this time period?
    Answer in units of m.

    3) Calculate the magnitude of the car’s average velocity from 5 s to 9 s.
    Answer in units of m/s.

    2. Relevant equations

    x - x0 = v0t + (1/2)at2

    3. The attempt at a solution

    For 1), I guess I'm confused at how to be reading the graph. I've tried breaking it up into 0 to 1 s, 1 to 3 s, 3 to 4 s, and 4 to 5 s, and using the equation above to find the displacement for each interval, but I'm getting confused as to where the acceleration is located on the graph. For example, is the acceleration at t = 3 s supposed to be 2 m/s2 or 0 m/s2?

    For 2), when it says that it doesn't experience any acceleration, does that mean acceleration = 0?

    For 3), what is the equation for average velocity?

    Any help/hints would be appreciated, thanks!
     
  2. jcsd
  3. Jul 24, 2011 #2

    gneill

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    Staff: Mentor

    Consider thinking in terms of open intervals (mathematically speaking) when the endpoints are discontinuous. So intervals might be:

    0 ≤ t < 1
    1 ≤ t < 3
    3 ≤ t < 4
    4 ≤ t < 5
    5 ≤ t < 6

    and so on.

    The end-point value that you use in your equations for any interval should reflect the value that the variable had within that interval immediately adjacent to the endpoint.
    Yes.
    Average velocity is defined to be the total displacement divided by the time.
     
  4. Jul 24, 2011 #3
    Hi gneill,

    So for 1), here's my work:

    0 to 1: x - 0 = 0(1) + .5(0)(1)^2 = 0
    1 to 3: x - 0 = 0(2) + .5(2)(2)^2 = 4
    3 to 4: x - 4 = 4(1) + .5(1)(1)^2 = 8.5
    4 to 5: x - 8.5 = 4(1) + .5(1)(1)^2 = 13

    Would 13 be the correct answer, or am I missing anything?
     
  5. Jul 24, 2011 #4

    gneill

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    Staff: Mentor

    Between 3 and 4 seconds the acceleration should be zero. So only the current velocity (from the end of the last stage) should be affecting the distance.

    You can move the initial position values to the RHS of your equations so that the equations make algebraic sense for the final equals sign. So for any given step:

    xnew = xp + vp*Δt + (1/2)*a*(Δt)2
    vnew = vp + a*Δt

    where the "p" subscript refers to the value at the end of the previous step.
     
  6. Jul 25, 2011 #5
    I've gotten everything except the last problem;

    3) Calculate the magnitude of the car’s average velocity from 5 s to 9 s.
    Answer in units of m/s.

    Here's my work so far:
    5 to 6: v = 13, x = 21.5
    6 to 8: v = 11, x = 45.5
    8 to 9: v = 3, x = 52.5

    You said the average velocity is total displacement divided by the time, so it would be:

    (21.5 + 45.5 + 52.5) / 4, right? But that answer is coming out as wrong...does anyone know what I'm doing wrong?

    I also tried (13 + 3) / 2, but that's wrong as well.
     
  7. Jul 25, 2011 #6

    gneill

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    Staff: Mentor

    You want the distance traveled during the specified time period, divided by the length of the time period.

    What is the displacement at the beginning of second 5? At the end of second 9?
     
  8. Jul 25, 2011 #7
    Oh...I think I understand...So it would be (21.5 + 52.5) / 4 ?
     
  9. Jul 25, 2011 #8

    gneill

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    Staff: Mentor

    Is 21.5 the displacement at the beginning of second 5 (same as end of second 4)?
     
  10. Jul 25, 2011 #9
    Ah, I read your previous post wrong, my fault... it would be 12.5.

    But would 52.5 still be the end point, right, since you can't extend the graph to t = 10?
     
  11. Jul 25, 2011 #10

    gneill

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    Staff: Mentor

    Correct.
     
  12. Jul 25, 2011 #11
    (12.5 + 52.5) / 4 is coming out to be incorrect...Any tips on what I'm doing wrong?
     
  13. Jul 25, 2011 #12

    gneill

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    Staff: Mentor

    Total displacement is the final position minus the initial position. Don't sum them!
     
  14. Jul 25, 2011 #13
    D'oh. Physics isn't my thing, but thanks for the help gneill - you've been invaluable in your patience and quick responses. Thank you!
     
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