## Graphs of Motion when *a *is constant but *a *≠ 0

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The graphs in the figure 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 position and velocity are initially \(200\text{ m}\) and \(15\text{ m/s},\) respectively.

The graph of position versus time in the (a) part of the figure above 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 position-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 (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 (b) is obtained. Furthermore, the slope of the graph of velocity versus time is acceleration, which is shown in figure (c).

## Example on 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 \(x\) vs. \(t\) 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 the figure above, where Q is the point at \(t = 25\text{ s}.\)

**Solution**

1. Find the tangent line to the curve at \(t = 25\text{ s}.\)

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, \(v\).

\(\text{slope} = v_{\text{Q}} = \cfrac{\Delta x_{\text{Q}}}{\Delta t_{\text{Q}}} = \cfrac{(3120\text{ m} \; – \; 1300\text{ m})}{(32\text{ s} \; – \; 19\text{ s})}\)

Thus,

\(v_{\text{Q}} = \cfrac{1820\text{ m}}{13\text{ s}} = 140\text{ m/s}.\)

**Discussion**

This is the value given in this figure’s table for \(v\) at \(t = 25\text{ s}\). The value of 140 m/s for \(v_{\text{Q}}\) is plotted in the figure above. The entire graph of \(v\) vs. \(t\) 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 \(v\) vs. \(t\) graph, rise = change in velocity \(\Delta v\) and run = change in time \(\Delta t\).

### The Slope of *v *vs. *t*

The slope of a graph of velocity \(v\) vs. time \(t\) is acceleration \(a\)*.*

\(\text{slope} = \cfrac{\Delta v}{\Delta t} = a\)

Since the velocity versus time graph in figure (b) is a straight line, its slope is the same everywhere, implying that acceleration is constant. Acceleration versus time is graphed in figure (c).

Additional general information can be obtained from the figure and the expression for a straight line, \(y = mx + b\).

In this case, the vertical axis \(y\) is \(V\), the intercept \(b\) is \(v_0\), the slope \(m\) is \(a\), and the horizontal axis \(x\) is \(t\). Substituting these symbols yields

\(v = v_0 + at.\)

A general relationship for velocity, acceleration, and time has again been obtained from a graph. Notice that this equation was also derived algebraically from other motion equations in the previous lessons.

It is not accidental that the same equations are obtained by graphical analysis as by algebraic techniques. In fact, an important way to *discover* physical relationships is to measure various physical quantities and then make graphs of one quantity against another to see if they are correlated in any way.

Correlations imply physical relationships and might be shown by smooth graphs such as those above. From such graphs, mathematical relationships can sometimes be postulated. Further experiments are then performed to determine the validity of the hypothesized relationships.