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When finding the acceleration using a ticker timer, how do i find the time taken?

  1. Aug 2, 2012 #1
    If i was running a vehicle down a slope with ticker tape attached to it, running through a ticker timer, and finding the distance of intervals of 5 (0-5, 6-10, 11-15, 16-20) and the time for each interval is 0.1, 0.2, 0.4 and 0.4.. If I already have the starting and final speed, how would I go about finding the time taken, in order for me to actually find the acceleration..Because the formula for acceleration is final speed-starting speed/time taken..

    I have this assessment due tomorrow, and im stuck on just this very thing.
    What is the time taken? Help me!

    Thanks
     
  2. jcsd
  3. Aug 2, 2012 #2

    CWatters

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    Perhaps try using this equation of motion and solving for "a"

    S=ut+0.5at2
     
  4. Aug 2, 2012 #3

    PeterO

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    Each time interval is 0.1 seconds long I believe, I think you meant that the time at the END of each interval was 0.1, 0.2 0.3 and 0.4.

    The average speed for the first time interval [0-5] is the length of tape, divided by the time interval [0.1] so numerically that works out at 10 x the length of the tape.
    he average speed for the first 0.1 seconds is generally taken as the instantaneous speed at time 0.05.

    Similar analysis of each other segment produces instantaneous speed at time 0.15, 0.25 and 0.35.

    By graphing those, you could calculate the acceleration [or perhaps take the gain in speed from 0.05 to 0.35, divided by the time interval [0.3].
     
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