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Prove This, please -- particle moving with a uniform acceleration

  1. Mar 3, 2015 #1
    1. The problem statement, all variables and given/known data

    Hi all. Please help me with this. A small particle moving with a uniform acceleration a covers a distances X and Y in the first two equal and consecutive intervals of time t. Show that a = Y-X/t².
    2. Relevant equations
    Acceleration is usually v-u/t i.e. rate of change of velocity with time.

    3. The attempt at a solution
    Since Y and X are distances, and t is time, and the unit for acceleration is m/s², I put two and two together. That was pretty easy, but I looked at the question properly and saw there was more to it than that, for instance, the 'first two equal and consecutive intervals of time t' I used different equations of motion such as s=ut+1/2at and v²=u²+2aS, but I kept getting stuck. Help, please!
     
  2. jcsd
  3. Mar 3, 2015 #2

    Dick

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    Suppose you start at x=0 and t=0 and accelerate uniformly at acceleration a. Then where are you at time t? Where are you at time 2t? Find X and Y from that and put them into (Y-X)/t^2.
     
  4. Mar 3, 2015 #3

    BvU

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    Hello VF, welcome to PF :smile:

    Please be meticulous about your equations. What you have to prove is that a = (Y-X)/t2.

    Your relevant equations should be like##\qquad## s=ut+1/2at2 (you forgot the square).
    And you need to explain what these symbols represent.

    And (if necessary) a second equation is v(t) = v0 + at

    But Dick (beat me to it) gives you a way to make do without needing this last equation.
     
  5. Mar 3, 2015 #4
    @BvU oops sorry. My smartphone keyboard is wonky(seriously), but that's beside the point.
    Yeah, I kind of did what @Dick said earlier, but notice that X and Y are different distances, and do not denote one point to another. The question says that an object will cover these two distances in the same time 2t with a uniform acceleration....
    Or is it saying so? I don't know.. I think I'm totally missing something. thanks for the tips, anyway! Really appreciate it.
     
  6. Mar 3, 2015 #5
    @BvU i'll be very sure to remember that next time. Thanks!
     
  7. Mar 3, 2015 #6

    Dick

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    If you tried to do what I suggested and still can't get it, it would help to post your work so we can see where you are going wrong. Maybe not on the wonky smartphone.
     
  8. Mar 4, 2015 #7

    BvU

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    My interpretation of
    • It starts at t=0 from rest at x = 0
    • at t = t1 it is at x = X
    • in the interval t from t = t1 to t2 = 2t1 it covers the distance from X to Y

    doesn't seem right. X and Y are different, yes. The remainder is unclear or incorrect.
     
  9. Mar 4, 2015 #8

    Dick

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    Agreed, but it doesn't have to start from rest for the formula to be valid.
     
  10. Mar 5, 2015 #9

    BvU

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    It's justified Dick corrects me. The exercise statement "in the first two equal and consecutive intervals of time t" lured me into interpreting as starting from v = 0. But a = (Y-X)/t2 also is true if the particle starts with v = v0.

    Now all VF has to do is come up with equations that feature these distances as functions of time and juggle with the given information.
    (the "relevant equation" in post #1 is not enough to solve this. There's nothing with the dimension length (distance) in there. You need more, e.g. from here)
     
  11. Mar 5, 2015 #10
    And.... SOLVED. Wow. That question really confused me... Thanks for all your help, @BvU and @Dick! :)
     
  12. Mar 5, 2015 #11
    My former physics teacher gave me a big hint I totally evaded.
     
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