We talk about Average speed and Average time, why not Average Distance? Ok, this might be a silly question, but nevertheless, this being quite fundamental, I would like a sound answer.
While solving arithmetic problems, we encounter problems where Average Speed or average time taken is asked, but why is average distance never asked?
What does average distance, if it means anything at all, mean? Why does it exist/not exist? 
Also, is Average time a valid question? 
I have some explanations in mind, but would love a good explanation with examples( both w.r.t Time speed distance problems and otherwise ) for clarity.
EDIT1: being more clear on the types of examples I'd require for understanding.
EDIT2: Regarding what I had in mind. How can we talk of averages when it comes to time and distance? They are absolute values unlike speed which is measured w.r.t time! Again, here I am considering 3 Dimensional systems(not including space). Which is why I say time is absolute.
 A: Probably because average distance is considered too easy.
I'm not entirely sure what you mean be an "average time problem" but the one thing that makes average speed problems different is that speed is not an accumulative measured value but a ratio between two accumulative values.  $\text{speed} = \dfrac {\text{distance}}{\text{time}}$
As a consequence the average speed over several trips is NOT the average of the various speeds.  $v_\text{average} \ne \frac {v_1 + v_2}2$.  That makes no sense if one trip is $60$ mph and the second trip is $120$ mph the average of the speed is $90$ mph but that doesn't "mean" anything.  What if the first trip was for only a few seconds and the second was for several hours.  Even if you normalize them so that the two trips last the same time, or the two trips are the same distance the answer is .... pointless.  What can we do or predict with the average of the speeds?  Nothing.
Average speed is the total distance divided by the total time and that is not the average of the speeds.  $$v_\text{average} = \frac {d_\text{total}}{t_\text{total}}= \frac {d_1+d_2}{t_1 + t_2} = \frac {v_1 t_1 + v_2 t_2}{t_1+t_2} \ne \frac{v_1 + v_2}2.$$
Distance on the other hand is linear and the average distance is the average of the distances.
$$d_\text{average} = \frac {d_1 + d_2}2.$$
Even if we express in terms of speed $\text{distance} = \text{speed} \times \text{time}$ and it distributes nicely.
$$d_\text{average} = \frac {d_1+d_2}2 = \frac{v_1 t_1 + v_2 t_2}{2} = \frac {v_\text{average}(t_2 + t_1)}{2} \text{ (because } = \frac {\frac {d_\text{total}}{t_\text{total}} (t_1 + t_2)}2=\frac {d_\text{total}}2) $$
Anyway, obviously we could have an average distant question.  Take so many trips each of a certain speed lasting a certain time, what is the average distance of each trip.  Straightforward to solve.
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Okay viewing the comments I guess average time, distance, and speed, are all okay.
Say tom does three trips.  1: is 60 mile at 45 mph for 1 hour 20 minutes, 2: is 75 miles at 60 mph for 1 hour and 15 minutes.  And 3: is 60 miles at 80 mph for 45 minutes.
Find the average distance, time, and speed pretending you don't know any of the distances, time and speed.
Average distance = $\frac {d_{total}}3 = \frac {v_1*t_1 + v_2*t_2+v_3*t_3}{3} = \frac{45*1\frac 13 + 60*1.25 + 80*.75}{3} = \frac {60+75 + 60}3 = 65$ miles.  It IS the average of the distances.
Average speed = $\frac{d_{total}}{t_{total}} = \frac {60+75+60}{1\frac 13 + 1.25 + .75} = \frac {195}{3 hours 20 minutes} = \frac {195}{3\frac 13} = 58.5$ mph.  It is NOT the average of the speeds.
Average time + $\frac{t_{total}}3 = \frac{d_1/v_1 + d_2/v_2 + d_3/v_3}3 = \frac{60/45 + 75/60+60/80}3 = \frac {3\frac 13}3 = 1\frac 19 = 1 hour 6 \frac 23 minutes$.  It IS the average of the times.
A: You can take an average of several distances, as with several times.
One place where "average time" occurs is in statistical things where, for example, a phone call arriving at a switchboard has no effect on the timing of another phone call arriving, and one speaks of the average time from the arrival of one phone call to the next.
"Average speed" arises when say a car passed a certain point at 1:00 PM and reached a point $35$ miles from there at 2:00 PM, so that you can say its average speed during that hour was 35 miles per hour.
Perhaps "average distance" does not occur in this way in standard examples in math textbooks, but there is no reason why you can't take an average of several distances. And astronomers do speak of such things as the average distance from the earth to the sun (since the distance undergoes seasonal variations).
A: There are many physical problems that concern some mean distance especially in statistical mechanics. This, about the mean distance between the molecules of a gas, is the more classical that I know.
A: As others have mentioned, there are several valid use cases for notions of average(or mean) distance. A mundane example would be:

If you alternately run $3$ and $5$km every day beginning from Monday
  through Saturday, what is the average distance you've run in the
  entire week?

Now, if you're worried about taking an average of time because it is absolute...that is strictly speaking, not true. But for all every day intents and purposes we can consider the absoluteness of time to be a given truth, and avoid questions like "What is time, really?".
The concept of arithmetic mean (which is what you are referring to as the "average"-the mean is actually one of many average quantities) is quite simple and independent of what you are measuring your quantity(for example, speed)  against. To see this more clearly, consider a $100m$ race amongst $10$ participants; you can find the average(mean) speed of the runners, and also the average(mean) time taken to complete the race.
In conclusion, in this context, the mean may be applied to any function of a single variable. What you obtain is a number that represents the equal distribution of quantity to all the data points you are working with. In the above example: $\text{(mean time)}\times\text{(number of runners)}=\text{(total time taken by all the runners)}$
