1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Quick question concerning conversions

  1. Sep 11, 2006 #1
    my book is rather vague on how to do unit conversions.


    I have 60 Mi/H and need to get it into M/s.


    Could i just set it up as:


    [(60 Mi/H)(1609m)] / 3600 s


    with this i come out to 26.8 m/s which sounds fairly correct.. am i making any errors?



    sorry.. i'm very confused about this!:frown:
     
  2. jcsd
  3. Sep 11, 2006 #2
    The easiest way, and the way usually taught, is to do each step seperately. So, you would have: [tex]\frac{60mi}{hr} * \frac{hr}{60min} * \frac{1min}{60sec} * \frac{1609meters}{1mi} = \frac{26.8meters}{sec} [/tex]

    The miles, hours, and minutes cancel out to give you meters/sec. You want to show how the units cancel out in your work, so in your calculation, you would want it to be 1609meters/mile and 1hr/3600seconds.
     
    Last edited: Sep 11, 2006
  4. Sep 11, 2006 #3

    Astronuc

    User Avatar

    Staff: Mentor

    The answer is correct, but it would be less confusing if one would work out the steps.

    60 mi/hr * (1609 m/mi) * (1 hr/3600 s) = 26.8 m/s. Show the ratios.

    Also, one could realize that 1 mi = 5280 ft, and 60 mi/hr (60 mph) is the same as 88 ft/sec. Now 88 ft/s * (1 m/ 3.28084 ft) = 26.8 m/s, or 88 ft/s * 0.3048 m/ft = 26.8 m/s
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?



Similar Discussions: Quick question concerning conversions
  1. Quick conversion ? (Replies: 6)

Loading...