Finding average velocity given curved graph? I know that the average velocity is given by the ratio
$$ \frac{\text{total distance}}{\text{total time}}
$$ It should be easy to figure out from a linear displacement-time graph, but what about when only part of it is curved? Here is the graph I have: 
I thought that the total distance on the left part is $(25-5) = 20$, and distance on the right part is $(25 - 10) = 15. 20 +15 = 35 = \text{total distance}$. Total time $= 8s$, so I thought avg velocity should be $35/8 - 4.375$, which is the wrong answer.
Where am I going wrong? How do I find total distance when there's a curved graph?
 A: Let $d(t)$ be the displacement function, in terms of $t$. The average velocity over the time span $0$ to $8$ seconds is given by the following ratio:
$$
\frac{d(8)-d(0)}{8-0}=\frac{10-5}{8-0}=\frac{5}{8}
$$
A: you are going wrong by thinking displacement, and distance are the same thing ( they aren't).  
edit:   displacement is the net distance traveled with direction, distance is simply distance.  example someone walks 3 feet north, 3 feet east, 3 feet south ,and 3 feet west, their net distance from where they started is 0 the direction could be null in that case, if they traveled it in say 12 seconds  the average speed would be 12 feet traveled / 12 seconds = 1 ft/s, the average velocity is 0 ft/s N ( lets just say that since direction is important in velocity and displacement).
A: Hint:
The total distance of the entire motion is $10-5=5 $.
Note that the velocity for the first $4$ second is opposite to the velocity on the other time of the motion.
A: The main issue with this question seems to be one of terminology.
We define the displacement of an object as a vector that points from that object's initial position to its current position. Displacement is a path independent quantity. The time derivative of the displacement is the velocity, and we can compute average velocity as the ratio of displacement and time. Velocity is also a vector.
We define distance or path length as the scalar length of the path. This quantity of course depends on the shape of the path as well as its endpoints. Its time derivative is called speed (also a scalar), and the average speed is the ratio of distance and time.
If you're asked to find the average velocity over a period of time, there's no need to worry about the details of the object's motion. Just look at the final position minus the initial position.
A: My approach is somewhat different. I would say that the average velocity is equal to the total distance traveled divided by the total time. To that you need to calculate
$$v_{avg}=\frac{1}{8}\int \sqrt{1+\left(\frac{dy}{dx} \right)^2}~dx$$
Since the curve portion is not specified, I assumed it's a power law of the form $y=5+x^n$ with $n=\log 20/\log 4$. At this point numerical methods are needed. I found that
$$v_{avg}\approx\frac{36.2983}{8}=4.5373$$
The other answers seem to suggest that the average velocity is path independent. Do you think that your average speed to go the market is independent of your route?
