This post will be about the relationship between distance, speed (or rate) and time, as well as the manipulation of units.

I think most of you are comfortable solving this problem:

If you are travelling at 60 km/hr, how far will you have gone after 2 hours? So if you travel 60 km every hour, after 2 hours you will have travelled 60 × 2 = 120 km. The 60 km/hr is a rate (or speed as normal people would call it) and the 2 hours is the time. The result of multiplying these two things gives a distance. So in equation form, the relationship between these three things is *d* = *rt*, where *d* is distance, *r* is the rate or speed, and *t* is time. Now this equation is in the form that allows solving for distance. But we could just as well use this equation to solve for an unknown rate or speed. If the problem was: you travel 120 km at a constant speed for 2 hours. How fast were you travelling? So the unknown thing here is rate. If I take the above equation and divide both sides by *t*, I get *r* = *d*/*t*. So for this problem, 120km/2hr = 60 km/hr. I can similarly solve for an unknown time.

Now notice how the units work out. in the original problem, I am multiplying a rate times a time. That is, the “hr” cancel to leave just “km” just as if they were variables:

\[\frac{\mathrm{km}}{\rlap{/}{\mathrm{h}}\rlap{/}{\mathrm{r}}}\hspace{0.33em}\times\hspace{0.33em}\rlap{/}{\mathrm{h}}\rlap{/}{\mathrm{r}}\hspace{0.33em}{=}\hspace{0.33em}{\mathrm{km}}

\]

In the second problem we are dividing a distance, km, by a time, hr. That is, km/hr and that is the unit of the answer, a rate. Now let’s do a problem where time is the unknown.

The speed of light is 299,792 km/sec. The average distance from the sun to the earth is 149,597,870 km. I say ‘average’ because the earth’s orbit around the sun is not perfectly circular. So how long does it take for a ray of light to travel from the sun to the earth? Or a more interesting (and morbid) way to ask this is, how long would it take before we knew that the sun exploded?

Going back to our rate equation and solving for *t* gives *t* = *d*/*r.* So

\frac{149,597,870\mathrm{km}}{299,792\mathrm{km}/\sec}\hspace{0.33em}\hspace{0.33em}{=}\hspace{0.33em}{499}\sec

\]

Again, looking at the units, and remembering that when dividing by a fraction, you get an equivalent multiplication problem by multiplying the numerator times the reciprocal of the denominator gives

\[\frac{\mathrm{km}}{\mathrm{km}/\sec}\hspace{0.33em}\hspace{0.33em}{=}\hspace{0.33em}\rlap{/}{\mathrm{k}}\rlap{/}{\mathrm{m}}\hspace{0.33em}\times\hspace{0.33em}\frac{\sec}{\rlap{/}{\mathrm{k}}\rlap{/}{\mathrm{m}}}\hspace{0.33em}{=}\hspace{0.33em}\sec

\]

So ‘sec’ is the appropriate unit for the answer. Let’s convert the answer to minutes. There are 60 sec/min so

\[\frac{{499}\hspace{0.33em}\sec}{{60}\hspace{0.33em}\sec{/}\min}\hspace{0.33em}{=}\hspace{0.33em}{499}\hspace{0.33em}\rlap{/}{s}\rlap{/}{e}\rlap{/}{c}\hspace{0.33em}\times\hspace{0.33em}\frac{\min}{60\rlap{/}{s}\rlap{/}{e}\rlap{/}{c}}\hspace{0.33em}{=}\hspace{0.33em}{8}{.}{32}\hspace{0.33em}\min

\]

So at any moment, you have a little more than 8 minutes to live. What a happy thought!