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Time and time intervals when using kinematic equations

  1. Nov 10, 2015 #1
    1. A space vehicle accelerates uniformly from 85 m/s at t=0s to 462 m/s at t=10s. What is avg acceleration? How far did the space vehicle move from t=2s to t=6s.

    2. Got the avg accel easily enough w basic equation for accel. Problem is conceptual with the displacement portion of the question.

    3. I got the correct answer by using x=xo + vit + 1/2at2 for t=2s and t=6s and then basically did Δx= x6s-x2s.

    BUT...I don't understand why I can't just use Δx=vit + 1/2at2 with t=4 because the time difference between 6s and 2s is 4s. Isn't the t in that equation essentially a Δt? Shouldn't putting 4 in there tell me the distance it will travel in 4 s?

    And why do I get a third different answer if I use: Δx=1/2(vi+vf)Δt?


    I seem to be confused about what Δx means and when t vs. Δt is used. Or something. Ugh. Sorry. Help!
     
  2. jcsd
  3. Nov 10, 2015 #2
    Can you tell me what are your results?
     
  4. Nov 10, 2015 #3

    haruspex

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    It will, but you need to plug in the right value for vi. What value did you use?
     
  5. Nov 10, 2015 #4
    Draw the graph and show us how you did it.
     
  6. Nov 10, 2015 #5

    gneill

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    You say you got the correct result for the distance covered between those times by taking the difference of the results. Using symbols for the two times, say ##t_a## and ##t_b## for the 2s and 6s times, write out the two expressions for the total distance using the formula. Take the difference of the two expressions and collect the terms. Can you spot the problem with just plugging in ##Δt = t_b - t_a## into your formula?
     
  7. Nov 10, 2015 #6
    Haruspex -- Thank you!!! THAT was it!! I was plugging in 85 m/s as the initial velocity, not calculating the initial velocity at 2 s and plugging that in. UGH!!!!! Smh. Thank you everyone for commenting and being willing to try to help!! :)
     
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