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Kinematics - getting different answers when using different units

  1. Oct 13, 2012 #1
    1. The problem statement, all variables and given/known data

    A jetliner touches down at 270 km/h. The plane then decelerates (i.e., undergoes acceleration directed opposite its velocity) at 4.5 m/s2. What’s the minimum runway length on which this aircraft can land?

    2. Relevant equations

    v2fx = v2ix + 2axΔx

    3. The attempt at a solution

    i got the solution doing this:

    (270km/hr)(1000m/km)(hr/3600s) = 75m/s

    0 = (75m/s)2 + 2(-4.5m/s2x
    (-9m/s)Δx = 5625m2/s2
    Δx = 625m

    which is the correct answer.

    however, i also tried to get the answer by converting everything to km/h, and i couldn't get the correct answer. i can't figure out why, and it's driving me crazy. here's what i tried:

    0 = (270km/hr)2 + 2(-4.5 m/s2x
    9m/s2Δx = 72900km2/hr2

    and then an aside for unit conversion:

    (9m/s2)(km/1000m)(3600s2/hr2) = 32.4km/hr2

    i plugged that in to get

    32.4km/hr2Δx = 72900km2/hr2
    Δx = 2250 km, which is much, much more than the 625 m that i got above.

    what am i doing wrong?!?

    thanks for the help, i really appreciate it
     
    Last edited: Oct 13, 2012
  2. jcsd
  3. Oct 13, 2012 #2
    The conversion of the units in acceleration is way off.
     
  4. Oct 13, 2012 #3
    how so?

    this is probably a silly question but i've reworked this like four times and don't see what i'm doing wrong...
     
  5. Oct 13, 2012 #4
    3600 times 3600 over 1000 must be obviously greater than 1000.
     
  6. Oct 13, 2012 #5

    PhysicoRaj

    User Avatar
    Gold Member

    You are wrong at converting -4.5 m/s2 to 32.4 km/hr2
    it is -58320 km/hr2

    4.5 m/s2=[4.5(10-3)]/(3600*3600)-1
    =(4.5*3600*3600)/1000
    =58320
    Hence,
    0=2702+2(-58320)Δx
    2702=116640*Δx
    72900=116640*Δx
    Δx=72900/116640
    Δx=0.625km ( which is 625m)
     
  7. Oct 13, 2012 #6
    thanks!
     
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