So the problem is, you have a car that's initially stationary, and you accelerate it at 10 m/s^2 for 10 s. At the end of 10 s its velocity is obviously going to be 10 m/s^2 * 10 s, or 100 m/s. The question is, how much distance does it cover during that time?
The non-calculus-based-physics answer is: well, the car started off stationary (0 m/s) and wound up at 100 m/s, so its average velocity (waves hands to distract students) was 50 m/s. So, we just multiply the average velocity by the time and, presto chango, the distance covered was 50 m/s * 10 s = 500 m.
To which Mari, unperturbed by the handwaving, asks, how do we know we can multiply the average of the car's starting and ending velocities by the time it was traveling? After all, it was going slower than 50 m/s for part of the journey and faster for part of the journey.
The answer is, you can't necessarily. The white lie the professor told was when he took the average of the car's starting and ending velocities and declared that to be the car's average speed. You can't figure out the average value of something by just looking at it at two instants. You have to look at it over the entire time you want to average it over.
So let's forget about average velocity for a moment and solve the problem using calculus. To solve the problem using calculus, the first thing we need to do is graph the car's velocity vs time. When you graph velocity vs time, the slope of your curve is acceleration. Since acceleration is constant for this problem, that means the curve is going to be a straight line. (When doing calculus, mathematicians tend to call all lines "curves", even the straight ones). The slope of our "curve" is 10 m/s^2.
Now, to figure out how much distance was covered, we need to do what calculus calls "integrating the curve". That basically means, draw two vertical lines, if necessary, to close off the shape between the curve and the X axis, and compute the area of that shape. In this case, we only need to draw one vertical line, on the right edge, because on the left edge the curve already touches the X axis. By the way, integrating gets more complicated if the curve dips below the X axis. Fortunately, it doesn't for this problem, so let's not worry about that complication right now.
So now we've got this picture:
The base of the triangle is the amount of time the car was traveling (10 s). The height of the right hand side is the final velocity (100 m/s). So to work out the area, we just do the old triangle area formula (Area = 1/2 * base * height) and get 1/2 * 10 s * 100 m/s = 500 m.
(You may be wondering why you can get distance by integrating the velocity curve. I can answer that question, but it will probably have to wait for another day).
Here's why the professor's "averaging trick" worked. If you go halfway up the triangle and make a horizontal cut, you can remove the top half of the triangle, turn it upside down, and fit it on top of the rest of the triangle.
The shapes only fit together because the acceleration is constant, so the lines are straight. Since all we did was rearrange parts of the triangle, the total area is the same. So if we figure out the area of this new shape, we'll still get the same answer, the distance the car traveled.
Area of a rectangle, of course, is just base * height. Base is still 10 s. The height of the rectangle, well, that's determined by where we cut the triangle. The place where we cut the triangle was halfway up (we had to cut it at the halfway point, because otherwise the two pieces wouldn't have fit together so nicely), so the height is 50 m/s. Multiply those together and you get 500 m.
Ok, now back to the question of average velocity. (a), how do we know that the average velocity of the car over the entire 10 s is the same as the number we get by averaging its starting and ending velocities? And (b), how do we know that we can multiply the average velocity of the car by the time to get the distance?
The short answer to both of these questions is that it all comes down to figuring out what we really mean by the "average velocity" of something whose velocity is constantly changing. To figure that out, you have to use calculus too, and it turns out that you have to use exactly the same calculus we just used--integrating the velocity curve. Only to work out the average velocity, you divide the area by the width you integrated over. For this problem, the area equals the distance and the width equals the time, so average velocity = distance / time, and therefore distance = average velocity * time.
Anyway, I hope that helps some, Mari. And others, I hope at least it was entertaining.