# Arizona State University| PHY 121: Univ Physics I: Mechanics|Homework 1-2| Average Velocity from a Position vs. Time Graph

Learning Goal: To learn to read a graph of position versus time and to calculate average velocity.

In this problem, you will determine the average velocity of a moving object from the graph of its position x(t) as a function of time t. A traveling object might move at different speeds and in different directions during an interval of time, but if we ask at what constant velocity the object would have to travel to achieve the same displacement over the given time interval, that is what we call the object’s average velocity. We will use the notation vave[t1,t2] to indicate average velocity over the time interval from t1 to t2. For instance, vave[1,3] is the average velocity over the time interval from t=1 to t=3.

### PART A. Consulting the graph shown in the figure, find the object’s average velocity over the time interval from 0 to 1 second.

Average velocity is defined as the constant velocity at which an object would have to travel to achieve a given displacement (difference between final and initial positions, which can be negative) over a given time interval, from the initial time ti to the final time tf. The average velocity is therefore equal to the displacement divided by the given time interval. In symbolic form, average velocity is given by

$v_{ave}\left[t_i,\:t_f\right]=\frac{x\left(t_f\right)-x\left(t_i\right)}{t_f-t_i}$

Since the final and initial positions are equal, the average velocity is 0 m/s.

### PART B. Find the average velocity over the time interval from 1 to 3 seconds.

$v_{ave}\left[t_i,\:t_f\right]=\frac{x\left(t_f\right)-x\left(t_i\right)}{t_f-t_i}=\frac{60-20}{3-1}=\frac{40}{2}=20\:m/s$

### PART C. Now find $v_{ave}\left[0,\:3\right]$.

$v_{ave}\left[0,\:3\right]=\frac{x\left(t_f\right)-x\left(t_i\right)}{t_f-t_i}=\frac{60-20}{3-0}=\frac{40}{3}=13.3\:m/s$

### PART D. Find the average velocity over the time interval from 3 to 6 seconds.

$v_{ave}\left[3,\:6\right]=\frac{x\left(t_f\right)-x\left(t_i\right)}{t_f-t_i}=\frac{20-60}{6-3}=\frac{-40}{3}=-13.3\:m/s$