Earth and Sun - fast question Guys is my logic correct? If Earth travels around the Sun in $29.78$ km/s and I have to calculate how long will it take to travel $1$km I did following:
$1s / 29.78 = 0.03358$ (to $5$ d.p)
I think it's ok but I would like to get confirmation on this.
Thank you.
 A: As J.M. suggests in the comments, just treat "km" and "s" as variables like $x$ and $y$.  Then calculate
$$
\frac{1 \operatorname{km}}{29.78 \operatorname{km/s}}
=\frac{1}{29.78} \frac{\operatorname{km}}{\operatorname{km/s}}
=\frac{1}{29.78} \frac{1}{\operatorname{1/s}}
=\frac{1}{29.78} \operatorname{s}
\approx 0.03358 \operatorname{s}
$$
Note that the answer came out with units of seconds, as it should have.  This does not guarantee that your calculation is correct — but if it hadn't come out with the right units, you'd have known that it wasn't correct.
For example, if you'd tried to calculate $\frac{1 \operatorname{s}}{29.78 \operatorname{km/s}}$, as in your question, you'd have ended up with an answer in units of $\operatorname{s^2/km}$ and known immediately that you'd calculated something other than you meant to.  (Of course, in this case the numerical value in front of the units would've been the same, but that's really just a coincidence.)
This technique of treating units as if they were unknown variables is called dimensional analysis, and it's very useful in many elementary physics problems — so much so, that in my experience most physics teachers will insist on it, and won't give you full marks on an answer if they can't see that you've handled the units correctly (instead of just guessing the correct unit and appending it to the numerical result).
Mind you, the really good ones will also insist that you explain the physics (and/or geometry) involved, to show that you understand the problem and aren't just blindly applying memorized or guessed formulae.  Even dimensional analysis doesn't quite substitute for actual understanding, although it does often go a long way. :-)
