An object's one-dimensional motion along the x -axis is...

An object's one-dimensional motion along the $x$ -axis is graphed in Fig. 2-1. Describe its motion. The velocity of the object at any instant is equal to the slope of the displacement-time graph at the point corresponding to that instant. Because the slope is zero from exactly $t=0 \mathrm{~s}$ to $t=2.0 \mathrm{~s}$, the object is standing still during this time interval. At $t=2.0 \mathrm{~s}$, the object begins to move in the $+x$ -direction with constant-velocity (the slope is positive and constant). For the interval $t=2.0 \mathrm{~s}$ to $t=$ $4.0 \mathrm{~s}$ $$ v_{a v}=\text { slope }=\frac{\text { rise }}{\text { run }}=\frac{x_{f}-x_{i}}{t_{f}-t_{i}}=\frac{3.0 \mathrm{~m}-0 \mathrm{~m}}{4.0 \mathrm{~s}-2.0 \mathrm{~s}}=\frac{3.0 \mathrm{~m}}{2.0 \mathrm{~s}}=1.5 \mathrm{~m} / \mathrm{s} $$ The average velocity is then $a v=1.5 \mathrm{~m} / \mathrm{s}$ - positive $X$ -direction. During the interval $t=4.0 \mathrm{~s}$ to $t=6.0 \mathrm{~s}$, the object is at rest; the slope of the graph is zero and $x$ does not change for that interval. From $t=6.0 \mathrm{~s}$ to $t=10 \mathrm{~s}$ and beyond, the object is moving in the $-x$ -direction; the slope and the velocity are negative. We have $$ v_{a v}=\text { slope }=\frac{x_{f}-x_{i}}{t_{f}-t_{i}}=\frac{-2.0 \mathrm{~m}-3.0 \mathrm{~m}}{10.0 \mathrm{~s}-6.0 \mathrm{~s}}=\frac{-5.0 \mathrm{~m}}{4.0 \mathrm{~s}}=-1.3 \mathrm{~m} / \mathrm{s} $$ The average velocity is then $a v=1.3 \mathrm{~m} / \mathrm{s}$ - negative $X$ -direction.

An object's one-dimensional motion along the $x$ -axis is graphed in Fig. 2-1. Describe its motion.
The velocity of the object at any instant is equal to the slope of the displacement-time graph at the point corresponding to that instant. Because the slope is zero from exactly $t=0 \mathrm{~s}$ to $t=2.0 \mathrm{~s}$, the object is standing still during this time interval. At $t=2.0 \mathrm{~s}$, the object begins to move in the $+x$ -direction with constant-velocity (the slope is positive and constant). For the interval $t=2.0 \mathrm{~s}$ to $t=$ $4.0 \mathrm{~s}$
$$
v_{a v}=\text { slope }=\frac{\text { rise }}{\text { run }}=\frac{x_{f}-x_{i}}{t_{f}-t_{i}}=\frac{3.0 \mathrm{~m}-0 \mathrm{~m}}{4.0 \mathrm{~s}-2.0 \mathrm{~s}}=\frac{3.0 \mathrm{~m}}{2.0 \mathrm{~s}}=1.5 \mathrm{~m} / \mathrm{s}
$$
The average velocity is then $a v=1.5 \mathrm{~m} / \mathrm{s}$ - positive $X$ -direction.
During the interval $t=4.0 \mathrm{~s}$ to $t=6.0 \mathrm{~s}$, the object is at rest; the slope of the graph is zero and $x$ does not change for that interval.
From $t=6.0 \mathrm{~s}$ to $t=10 \mathrm{~s}$ and beyond, the object is moving in the $-x$ -direction; the slope and the velocity are negative. We have
$$
v_{a v}=\text { slope }=\frac{x_{f}-x_{i}}{t_{f}-t_{i}}=\frac{-2.0 \mathrm{~m}-3.0 \mathrm{~m}}{10.0 \mathrm{~s}-6.0 \mathrm{~s}}=\frac{-5.0 \mathrm{~m}}{4.0 \mathrm{~s}}=-1.3 \mathrm{~m} / \mathrm{s}
$$
The average velocity is then $a v=1.3 \mathrm{~m} / \mathrm{s}$ - negative $X$ -direction.

Key Concepts

Intervals of Rest and Constant Motion

Analyzing different time intervals on the displacement-time graph allows one to distinguish between periods when an object is at rest, moving with constant velocity, or moving with acceleration. Recognizing these intervals helps in understanding the behavior of the motion, such as identifying segments of inactivity or directional changes, which is essential in a comprehensive study of kinematics.

Average Velocity

Average velocity is defined as the total displacement divided by the total time taken for that displacement. It is represented by the slope of a secant line connecting two points on the displacement-time graph. This concept is crucial for quantifying motion over a period when the velocity might not be constant, allowing one to summarize the overall effect of varied motion.

Displacement-Time Graph

A displacement-time graph displays the position of an object along a given axis as a function of time. Such graphs are fundamental in kinematics as they visually represent how an object’s position changes, allowing one to infer details about its motion, such as periods of rest, constant motion, or acceleration, simply by examining the slope of the graph at any given point.

Slope as Velocity

The slope of a displacement-time graph at any point represents the instantaneous velocity of the object. A positive slope indicates motion in the positive direction, a negative slope indicates motion in the negative direction, and a zero slope implies that the object is at rest. This relationship is key for understanding and analyzing motion in physics, as it connects graphical representations with physical quantities.

Transcript

00:03 Today i'm going to be looking at the relationship between position, time, and motion.

00:09 So position, time, and philosophy is another way to say it.

00:12 And what i have is a graph of position versus time, and i want to use that graph to find out some things about the motion of the object that this graph is representing.

00:23 So let me draw out the graph for you first.

00:27 All right.

00:31 So this is my origin zero, zero.

00:35 The x -axis is going to be my time and the y axis.

00:40 So i'll just call that my y, my position in centimeters.

00:47 All right, let me just draw some 2, 4, 2, 4, 2, 4, 2, 4, 2, 4, 6, 10.

01:07 So our graph, we have our, from time equals zero to two seconds, we have no changing position.

01:18 Then at two to four seconds, we move up to a position of y equals three, once three centimeters.

01:30 And then from four to six seconds, we again have no change in position.

01:38 And then at six seconds, we're going to six seconds.

01:40 Are going to move in a negative direction to negative 2 centimeters at 10 seconds.

01:56 So let's look at this in four parts.