A question about converting dimensions

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To convert acceleration from miles per hour times second (mi/h*s) to meters per second squared (m/s^2), the discussion emphasizes using conversion factors for both distance and time. Specifically, miles must be converted to meters using the factor of approximately 0.0006214 mi/m, and hours to seconds using 3600 sec/hr. The correct setup involves multiplying these conversion factors while ensuring units cancel appropriately. A discrepancy arises with a book answer of 0.427 m/s^2, which participants suggest may not match their calculations. The conversation highlights the importance of careful unit conversion and verification of results.
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My question seems rather simple, but I could't have the right answer. Hope you can help me about it. Thanks!



Determine a convertion factor that, upon multiplication, changes an acceleration in miles per hour times second (mi/h*s) to meter per second squared (m/s^2).
 
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To arrive at your conversion factors, consider the units that need to be changed one by one. Firstly, the unit for measuring length. We need to change miles to metres. How many miles per metre? My computer tells me 6.214 * 10^(-4) mi/m

Do this for the units of time as well. How many seconds per hour? (60 sec/min)*(60 min/hr) = 60^2 sec/hr = 3600 sec/hr.

Now, set things up so that you successively multiply by the required conversion factors in a chain, thereby being easily able to keep track of how the old units cancel out leaving only the new ones. The conversion factors are fractions, and so in order to know which way around they should be written, you need to keep track of whether the unit that you are trying to cancel in the original expression is on top or on the bottom:

1\ \ \frac{\textrm{mi}}{\textrm{h} \cdot \textrm{s}} \times \frac{1\ \ \textrm{m}}{0.0006214\ \ \textrm{mi}} \times \frac{1\ \ \textrm{h}}{3600\ \ \textrm{s}} =

continued in next post (trying to sort out my LaTeX)
 
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Welcome to PF!

MrEnergy said:
Determine a convertion factor that, upon multiplication, changes an acceleration in miles per hour times second (mi/h*s) to meter per second squared (m/s^2).

Hi MrEnergy! Welcome to PF! :smile:

Show us what factors you're using …

you've probably made a simple mistake that we can spot for you. :smile:
 
cepheid said:
1 \frac{\textrm{mi}}{\textrm{h} \cdot \textrm{s}} \times \frac{1\, \textrm{m}}{0.0006214\, \textrm{mi}} \times \frac{1\, \textrm{h}}{3600\, \textrm{s}} =

continued in next post (trying to sort out my LaTeX)

= \frac{1}{0.0006214 \times 3600} \, \, \, \frac{\textrm{mi} \cdot \textrm{m} \cdot \textrm{h}}{\textrm{h} \cdot \textrm{s} \cdot \textrm{mi} \cdot \textrm{s}}

Now cancel out the units that are in both numerator and denominator, and you should be left with what you were trying to get (a good way to check if you did it right!). Until you gain more confidence, this is the most transparent and foolproof way to set up unit conversions. Note that multiplying by your conversion factors amounts to little more than clever ways of multiplying by 1.
 
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Firstly thank you for replying, secondly, i also did the same thing that you did, but the answer given on the book was different to the one that I've found...That's why I've asked..

The book gave an answer like =0,427(m/s^2)(mi/h*s) but that equation didn't give the exact... Any ideas??
 
Multiply out cepheid's numbers from post #4. You can something close to, but not exactly, the 0,427 figure you give.
 
Ok, at least now I think that I did right. And yeah I don't get the 0,427 number...
 
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