Slopes and General Relationships

Graphs typically have two perpendicular axes: a horizontal axis (usually the independent variable) and a vertical axis (usually the dependent variable).

If the horizontal axis is called [latex]x[/latex] and the vertical axis called [latex]y[/latex], a straight-line graph as in Figure 15.1 can be written as:

[latex]y = mx + b[/latex]

where:

• [latex]m[/latex] is the slope (rise over run)

• [latex]b[/latex] is the y-intercept (where the line crosses the vertical axis)

.

Graph of Displacement vs. Time (a = 0, so v is constant)

Time is usually an independent variable, so a graph of displacement ([latex]x[/latex]) versus time ([latex]t[/latex]) has [latex]t[/latex] on the horizontal axis and [latex]x[/latex] on the vertical axis.

Figure 15.2 shows such a graph for a jet-powered car on a flat dry lake bed.

Using the straight-line formula, the slope of this graph represents the average velocity [latex]\bar{v}[/latex], and the y-intercept corresponds to the initial displacement [latex]x_0[/latex].

Substituting these into the straight-line equation gives:

[latex]x = \bar{v} t + x_0[/latex]

or equivalently:

[latex]x = x_0 + \bar{v} t[/latex]

Thus, a graph of displacement vs. time expresses the general relationship between displacement, velocity, and time while giving detailed numerical information about a specific situation.

Graphs of Motion when [latex]a[/latex] is constant but [latex]a\ne 0[/latex]

The graphs in Figure 15.3 below represent the motion of the jet-powered car as it accelerates toward its top speed, but only during the time when its acceleration is constant. Time starts at zero for this motion (as if measured with a stopwatch), and the displacement and velocity are initially 200 m and 15 m/s, respectively.

The graph of displacement versus time in Figure 15.3(a) is a curve rather than a straight line. The slope of the curve becomes steeper as time progresses, showing that the velocity is increasing over time. The slope at any point on a displacement-versus-time graph is the instantaneous velocity at that point. It is found by drawing a straight line tangent to the curve at the point of interest and taking the slope of this straight line. Tangent lines are shown for two points in Figure 15.3(a). If this is done at every point on the curve and the values are plotted against time, then the graph of velocity versus time shown in Figure 15.3(b) is obtained. Furthermore, the slope of the graph of velocity versus time is acceleration, which is shown in Figure 15.3(c).

Example 15.2: Determining Instantaneous Velocity from the Slope at a Point: Jet Car

Calculate the velocity of the jet car at a time of 25 s by finding the slope of the [latex]x[/latex] vs. [latex]t[/latex] graph in the graph below.

Strategy

The slope of a curve at a point is equal to the slope of a straight line tangent to the curve at that point. This principle is illustrated in Figure 15.5, where Q is the point at [latex]t=\text{25 s}[/latex].

Solution

1. Find the tangent line to the curve at [latex]t=\text{25 s}[/latex].
2. Determine the endpoints of the tangent. These correspond to a position of 1300 m at time 19 s and a position of 3120 m at time 32 s.
3. Plug these endpoints into the equation to solve for the slope, [latex]v[/latex].
   
   [latex]\text{slope}={v}_{Q}=\frac{{\Delta x}_{Q}}{{\Delta t}_{Q}}=\frac{\left(\text{3120 m}-\text{1300 m}\right)}{\left(\text{32 s}-\text{19 s}\right)}[/latex]

   Thus,

   [latex]{v}_{Q}=\frac{\text{1820 m}}{\text{13 s}}=\text{140 m/s.}[/latex]

Discussion

This is the value given in this figure’s table for [latex]v[/latex] at [latex]t=\text{25 s}[/latex]. The value of 140 m/s for [latex]{v}_{Q}[/latex] is plotted in Figure 15.5. The entire graph of [latex]v[/latex] vs. [latex]t[/latex] can be obtained in this fashion.

Carrying this one step further, we note that the slope of a velocity versus time graph is acceleration. Slope is rise divided by run; on a [latex]v[/latex] vs. [latex]t[/latex] graph, rise = change in velocity [latex]\Delta v[/latex] and run = change in time [latex]\Delta t[/latex].

Graphs of Motion Where Acceleration is Not Constant

Consider the motion of the jet car accelerating from 165 m/s to its top speed of 250 m/s, graphed in Figure 15.6. The time starts at zero, initial displacement is 2900 m, and initial velocity is 165 m/s (final values from the motion in Figure 15.3). The acceleration gradually decreases from

[latex]5.0\ \text{m/s}^2[/latex]

to zero as the car reaches 250 m/s. The slope of the displacement vs. time graph increases until [latex]t=55\ \text{s}[/latex], after which it remains constant. The velocity increases until 55 s and then remains constant because acceleration drops to zero (see Figure 15.6).

Graphs of displacement vs. time can be used to generate velocity vs. time graphs, and velocity vs. time graphs can generate acceleration vs. time graphs by calculating the slope at each point. Linear graphs have constant slopes, simplifying this process.

Check Your Understanding

A graph of velocity vs. time for a ship entering a harbor is shown below.

• (a) Describe the ship’s motion based on the graph.

• (b) Sketch the ship’s acceleration vs. time graph.

1. The ship moves at constant velocity and then begins to decelerate at a constant rate. At some point, its deceleration rate decreases. It maintains this lower deceleration rate until it stops moving.
2. A graph of acceleration vs. time would show zero acceleration in the first leg, large and constant negative acceleration in the second leg, and constant negative acceleration.