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Homework Help: Position vs Time problem

  1. Sep 29, 2008 #1
    1. The problem statement, all variables and given/known data

    The position versus time for a certain particle moving along the x axis is shown in the figure below.

    http://img87.imageshack.us/img87/6101/physicswg4.th.jpg [Broken]http://g.imageshack.us/thpix.php [Broken]

    Find the average velocity in the following time intervals.

    (a) 0 to 4 s
    _____ m/s

    (b) 0 to 5 s
    _____ m/s

    (c) 4 s to 5 s
    _____ m/s

    (d) 5 s to 7 s
    ______ m/s

    (e) 0 to 8 s
    ______ m/s


    2. Relevant equations

    I thought the equation to use would be x(t_2) - x(t_1) / t_2 - t_1

    3. The attempt at a solution

    After following that equation, I got 5 m/s for (a), but it was incorrect.

    Could someone PLEASE tell me what I did wrong. Thanks.
     
    Last edited by a moderator: May 3, 2017
  2. jcsd
  3. Sep 29, 2008 #2

    alphysicist

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    Hi chanv1,

    Your equation looks right to me; but I am not getting 5m/s for part a. What numbers are you plugging into your equation (for x(t2), x(t1), t2, and t1) to get 5m/s?
     
  4. Sep 29, 2008 #3
    I did

    5(4) - 0(0) / 4-0 = 5

    What answer did you come up with? and would you please show me how?
     
  5. Sep 29, 2008 #4

    alphysicist

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    I think you are visualizing the formula incorrectly. It is:

    [tex]
    v_{\rm ave} = \frac{x(t_2)-x(t_1)}{t_2-t_1}
    [/tex]
    and [itex]x(t_2)[/itex] is not x times [itex]t_2[/itex]. So it might be better to write it as:

    [tex]
    v_{\rm ave} = \frac{x_2-x_1}{t_2-t_1}
    [/tex]

    because [itex]x(t_2)[/itex] means the position x at time [itex]t_2[/itex], so [itex]x(t_2)=x(4 \mbox{ seconds}) = 5\mbox{ meters}[/itex]
     
  6. Sep 29, 2008 #5
    Yeah, I know. I don't understand what I'm doing wrong ... please walk me through this?

    Isn't the distance for 4s, 5? so I would 5 * 4 = 20 and so on?

    What should I be seeing instead?
     
  7. Sep 29, 2008 #6

    alphysicist

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    I seem to be having some computer problems, so I'll repost the edits I made in my last post in case they just are not showing up.

    it might be better to write the average velocity formula as:

    [tex]
    v_{\rm ave} = \frac{x_2-x_1}{t_2-t_1}
    [/tex]

    because [itex]x(t_2)[/itex] means the position x at time [itex]t_2[/itex], so [itex]x(t_2)=x(4 \mbox{ seconds}) = 5\mbox{ meters}[/itex], not 20.
     
  8. Sep 29, 2008 #7
    so would the answer then be 1.25 or 1.3?

    but that answer was incorrect too.
     
  9. Sep 29, 2008 #8

    alphysicist

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    Unless I'm just not seeing something, it looks like 1.25m/s is the correct answer to me. (Since you mentioned 1.3m/s, did you try 1.25 or did you input the rounded answer?)
     
  10. Sep 29, 2008 #9
    yeah, I rounded the number. Silly me!
    Thanks for all your help!
     
  11. Sep 29, 2008 #10

    alphysicist

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    Sure, glad to help!
     
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