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What is a distance an object went through at the speed of 14hm/min for 6 hours?

I am having problems with conversions.

First, I tried passing 14hm/min to m/s. I did it alright. But now I am unsure. Do I multiply it by 6 hours and I'm done?

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It might be sensible to turn 6 hours into minutes or seconds before you multiply by the speed so that the time units are the same. –  Henry May 13 '11 at 7:34
    
@Omega - Is there a typo in your unit for the speed? 14hm/min doesn't make any sense.. –  qftme May 13 '11 at 9:23
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That "h" might stand for "hekt(o)-" (= x100 ) ? Typical homework! In practice the hect- is used for ha (10 000 m²) and nowadays hecto-Pascal, nothing else. –  Georg May 13 '11 at 9:28
    
@qftme: It does make sense if $hm$ stands for hectometer, cf. en.wikipedia.org/wiki/Hectometre –  Qmechanic May 13 '11 at 9:30
    
@Georg and @Qmechanic An unusual unit for distance. If that is indeed the case here, well .. fine. –  qftme May 13 '11 at 9:34
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3 Answers 3

up vote 2 down vote accepted

You can leave the 14hm/min just as it is, if you like. Convert the 6 hours to how many minutes there are in 6 hours, 6 hours times 60 minutes per hour = 360 minutes. The comments on your Question are beside the point, hm could be half-meters, hectometers, whatever, ... — if one travels 14 “hm”s in 1 minute, then one travels 360 times 14 “hm”s in 360 minutes.

You could also convert the 14 hm/min into hm/hour, $60\times 14$ hm/hour, then multiply that by the number of hours. It's possible to keep track of the units quite nicely by arranging everything on two lines, $$\frac{14\ {\rm hm}}{{\rm minute}}\times\frac{60\ {\rm minute}}{{\rm hour}}\times 6\ {\rm hour} = 14\times 60\times 6 {\rm hm}.$$ The different units all cancel, leaving behind the conversion factors. If you do this methodically every time then this kind of conversion, and harder ones, will become second nature.

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The TeX wasn't coming out readable, at least in my browser. I edited to fix. I hope I didn't screw it up. –  Ted Bunn May 13 '11 at 14:11
    
@Ted It came out alright while I was editing it, but then I think it didn't just now. Perhaps I went a little too far. Sometimes I almost enjoy writing out the TeX more than the content -- which is probably a character flaw as far as being a Physicist goes. So thanks for the corrective! Is there a 12-step program? –  Peter Morgan May 13 '11 at 14:18
    
not only was the question very close to being a "do my homework" type question anyway, clearly not permitted in the FAQ, but your answer has gone one step further and does it. Hence why I phrased my attempt more as a guide than complete calculation. –  qftme May 13 '11 at 14:20
    
I have an unhealthy enjoyment of monkeying with TeX too. –  Ted Bunn May 13 '11 at 14:23
    
@qftme of course, I could see the impulse not to do the Question in your Answer, but the Question seemed to me to show that the person wasn't "getting it", and it seemed -- again, to me -- that your Answer didn't help much with that. My Answer is geared to helping them get it, I hope. I somewhat regretted that I gave a complete Answer, but I hope Omega doesn't have only this one question on unit conversions to do. Being well organized will, almost always, be worthwhile. –  Peter Morgan May 13 '11 at 14:37
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Peter Morgan's answer is the most straightforward. It doesn't matter what an 'hm' is. If you go 14 of them in one minute, then you go:

(14hm/min) * (60 min/hr) * (6 hr) = (14 * 360) hm = 5040 hm in 6 hours.

The technique you need to master, in unit conversion, is unit cancellation. Notice, in the equation above, that 'min' in the numerator of the first term cancels out 'min' in the denominator in the second. Likewise for 'hr' in the second and third terms. All that is left is 'hm' and a little math.

Now, if you want to convert 'hm' to meters, you'll need to tell us what an 'hm' is.

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The answer to this question is yes, but... Remember that when we multiple quantities together it is important that they are all in standard (SI) units. This will make it easier to keep track of what units your answer will be in.

The question is one that concerns speed ($s$), distance ($d$) and time ($t$). The equation relating them is:

$$ s=d/t$$

Rearranging this to get the distance as the subject (as per your question) gives:

$$ d=st$$

Like I said- once you've converted all the values to SI units, it's merely a case of putting the values into your calculator and seeing what you get.

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