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(Easy) Finding initial velocity from uniform acceleration

  1. Feb 6, 2013 #1
    Hi, I don't know what I'm doing wrong but these significant figures seem to be screwing me up.
    Problem:
    A bike rider accelerates constantly to a velocity of 7.5 m/s during 4.5 s. The bike's displacement is +19 m. What was the initial velocity of the bike?
    The textbook says the answer is 0.94 m/s, but I got 0.9 m/s. I've attached an image showing the steps I took to get to my answer...where did I go wrong?
    ORFsRC9.jpg
     
  2. jcsd
  3. Feb 6, 2013 #2
    Your working is correct, What you can also do is set up two simultaneous equations. your given enough information to set up two kinematic equations with two unkowns, (unknowns should be acceleration and initial velocty) then solve simultaneously. both ways get same answer and is correct its just that you have missed out a decimal place in your answer which should give you 0.94 as you said. remember answer must have the same number of significant figures as the number with the most sig figs in the question.
     
  4. Feb 6, 2013 #3
    try to work to as many decimal places as you can in your working out then cut down to the correct significant figures in your final answer
     
  5. Feb 6, 2013 #4
    Oh okay, thank you. I did not know that my answer is supposed to have the same number of significant figures as the number with the most sig figs in the question. So I work it out to as many decimal places as I can, then cut it down to the number with the most sig figs? In future cases, does this still apply if the number with the most sig figs is of a different measurement (time, displacement, velocity, etc) than my answer?
     
  6. Feb 6, 2013 #5

    tiny-tim

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    welcome to pf!

    hi 2k8bomb! welcome to pf! :smile:

    no, the answer should be the least number of significant figures in the question

    and, most importantly, do not round off to that number during the calculation!! :rolleyes:

    (you can round off a little, but not completely)

    in this case, your 38/4.5 should have been 8.444 (not 8.4), so that when you subtracted 7.5 you still had 3 sig figs (0.944) that you could finally round off correctly (up or down) to 2 sig figs (0.94) :wink:

    see https://www.physicsforums.com/library.php?do=view_item&itemid=523

    EDIT: CORRECTION, see below :redface:
     
    Last edited: Feb 6, 2013
  7. Feb 6, 2013 #6
    sorry 2k8bomb i meant least amount, always get that mixed up....
     
  8. Feb 6, 2013 #7

    tiny-tim

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    actually, on second thoughts :redface:, i think your answer (0.9) was correct

    my warning not to round off completely during the caclulation is still correct, but its only purpose is to avoid cumulative rounding errors (eg, 2.4 + 2.4 ends up as 2 + 2 = 4 instead of 5 ! :rolleyes:)

    the problem here was not cumulative rounding, it was that there was a multiplication followed by an addition (or subtraction), for which the rules are different

    this this case, you had 8.444 minus 7.5 …

    the rule for addition/subtraction (as i suspect you know :wink:) is to use the smallest number of decimal places, in this case 1 place, giving 0.9

    (compare the three closest available inputs, 7.4 7.5 and 7.6 … the results are 0.844 0.944 and 1.044, which obviously are only reliable to 1 decimal place: 0.8 0.9 and 1.0)

    so i think the book is wrong

    what do other people think? :smile:
     
  9. Feb 6, 2013 #8
    Thank you for the response, but in reference to sig figs, isn't 38 divided by 4.5 = 8.4? Since they both have 2 significant figures. Argh, this should be the easy part of physics but I find myself getting easily confused :tongue:
     
  10. Feb 6, 2013 #9

    tiny-tim

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    hi 2k8bomb! :smile:
    if that's the end of the question, yes, you do it to 2 sig figs, and 38/4.5 = 8.4

    but it's only an intermediate step, so you must always keep at least one extra sig fig, to be safe

    (in this case, it makes no difference to the final result, but it's best to be safe)

    suppose instead of 38/4.5 - 7.5 it was 5.1/4.5 - 0.77

    that would be 1.13 - 0.77 = 0.36 to 2 decimal places

    (if you round off too early, you get 1.1 - 0.77 = 0.33 :wink:)​
     
  11. Feb 6, 2013 #10
    Great, thank you. That cleared up a bunch of past confusion and potential future frustration as well.
     
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