# Which is the proper way to calculate average speed.

Let's say we have a graph, a distance-time graph, and there are couple points on the graph. The points are connected to each other with different slopes, so with different speeds between the points.

What is the correct way of calculationg speed?

Method 1

Subtract the initial distance from the final distance from the whole graph and divide by time.

Method 2

Subtract the initial distance from final distance of each segment of speed. Then divide each segment difference by time, abd average them all together.

I keep insisting on method 2 and my teacher on method 1. Which is more correct? By the way they produce similar but different results.

• Your method is incorrect unless you properly weight the average. Going 20 mph for one hour will influence the average speed more than going 100 mph for 5 minutes. So for instance, in that example the average would be: $\frac{60\cdot20 + 5\cdot 100}{65} \approx 26$. – Jared Jun 17 '16 at 5:29
• Also, do you mean average speed or average velocity? There is a difference. Method 1 correctly computes the average velocity but not the average speed. – Jared Jun 17 '16 at 5:31
• Isn't velocity same thing as speed, only difference being velocity has direction, which really doesnt matter? – KKZiomek Jun 17 '16 at 5:33
• Say a dog was running around a circular track with length 10 meters, and the dog took 20 seconds to come back to the start line, so that the final position is equal to the initial position. Then the average velocity would be 0, since (final position)-(initial position)=0, but the average speed would be 0.5 m/s, since speed is (total distance traveled)/(time taken). – zxcvber Jun 17 '16 at 5:39
• Assuming this actually is a "distance"-time graph then the teacher's method would be correct. Because if it is a distance-time graph then the graph must be a monotonically non-decreasing function. On the other hand, if it's a displacement-time graph then you have to take into account regions where the graph "turns around" (i.e. decreases). – Jared Jun 17 '16 at 5:56

For a function $f(x)$, its average value on an interval $[a, b]$ is given as; $$f_{avg} =\frac{1}{b-a}\int_a^b f(x)dx$$
Suppose the given distance function respect to time ($t$) is given by $s(t)$. Then the velocity would be given as $v(t)=s'(t)$
Now if we were to calculate the average velocity, we would compute; $$v_{avg}=\frac{1}{b-a}\int_a^b v(t)dt$$
Using the fundamental theorem of calculus, $$v_{avg}=\frac{1}{b-a}\int_a^b v(t)dt=\frac{1}{b-a}[s(t)]_a^b$$
which is equal to $$\frac{s(b)-s(a)}{b-a}$$