# 2.8 Graphical analysis of one-dimensional motion  (Page 9/8)

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• Describe a straight-line graph in terms of its slope and y -intercept.
• Determine average velocity or instantaneous velocity from a graph of position vs. time.
• Determine average or instantaneous acceleration from a graph of velocity vs. time.
• Derive a graph of velocity vs. time from a graph of position vs. time.
• Derive a graph of acceleration vs. time from a graph of velocity vs. time.

A graph, like a picture, is worth a thousand words. Graphs not only contain numerical information; they also reveal relationships between physical quantities. This section uses graphs of displacement, velocity, and acceleration versus time to illustrate one-dimensional kinematics.

## Slopes and general relationships

First note that graphs in this text have perpendicular axes, one horizontal and the other vertical. When two physical quantities are plotted against one another in such a graph, the horizontal axis is usually considered to be an independent variable    and the vertical axis a dependent variable    . If we call the horizontal axis the $x$ -axis and the vertical axis the $y$ -axis, as in [link] , a straight-line graph has the general form

$y=\text{mx}+b.$

Here $m$ is the slope    , defined to be the rise divided by the run (as seen in the figure) of the straight line. The letter $b$ is used for the y -intercept , which is the point at which the line crosses the vertical axis.

## Graph of displacement vs. time ( a = 0, so v Is constant)

Time is usually an independent variable that other quantities, such as displacement, depend upon. A graph of displacement versus time would, thus, have $x$ on the vertical axis and $t$ on the horizontal axis. [link] is just such a straight-line graph. It shows a graph of displacement versus time for a jet-powered car on a very flat dry lake bed in Nevada.

Using the relationship between dependent and independent variables, we see that the slope in the graph above is average velocity $\stackrel{-}{v}$ and the intercept is displacement at time zero—that is, ${x}_{0}$ . Substituting these symbols into $y=\text{mx}+b$ gives

$x=\stackrel{-}{v}t+{x}_{0}$

or

$x={x}_{0}+\stackrel{-}{v}t.$

Thus a graph of displacement versus time gives a general relationship among displacement, velocity, and time, as well as giving detailed numerical information about a specific situation.

## The slope of x Vs. t

The slope of the graph of displacement $x$ vs. time $t$ is velocity $v$ .

$\text{slope}=\frac{\Delta x}{\Delta t}=v$

Notice that this equation is the same as that derived algebraically from other motion equations in Motion Equations for Constant Acceleration in One Dimension .

From the figure we can see that the car has a displacement of 25 m at 0.50 s and 2000 m at 6.40 s. Its displacement at other times can be read from the graph; furthermore, information about its velocity and acceleration can also be obtained from the graph.

## Determining average velocity from a graph of displacement versus time: jet car

Find the average velocity of the car whose position is graphed in [link] .

Strategy

The slope of a graph of $x$ vs. $t$ is average velocity, since slope equals rise over run. In this case, rise = change in displacement and run = change in time, so that

$\text{slope}=\frac{\Delta x}{\Delta t}=\stackrel{-}{v}.$

Since the slope is constant here, any two points on the graph can be used to find the slope. (Generally speaking, it is most accurate to use two widely separated points on the straight line. This is because any error in reading data from the graph is proportionally smaller if the interval is larger.)

Solution

1. Choose two points on the line. In this case, we choose the points labeled on the graph: (6.4 s, 2000 m) and (0.50 s, 525 m). (Note, however, that you could choose any two points.)

2. Substitute the $x$ and $t$ values of the chosen points into the equation. Remember in calculating change $\left(\Delta \right)$ we always use final value minus initial value.

$\stackrel{-}{v}=\frac{\Delta x}{\Delta t}=\frac{\text{2000 m}-\text{525 m}}{6\text{.}\text{4 s}-0\text{.}\text{50 s}},$

yielding

$\stackrel{-}{v}=\text{250 m/s}.$

Discussion

This is an impressively large land speed (900 km/h, or about 560 mi/h): much greater than the typical highway speed limit of 60 mi/h (27 m/s or 96 km/h), but considerably shy of the record of 343 m/s (1234 km/h or 766 mi/h) set in 1997.

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