# 2.8 Graphical analysis of one dimensional motion  (Page 2/8)

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## Graphs of motion when $a$ Is constant but $a\ne 0$

The graphs in [link] 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 [link] (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 [link] (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 [link] (b) is obtained. Furthermore, the slope of the graph of velocity versus time is acceleration, which is shown in [link] (c).

## 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 [link] , where Q is the point at $t=\text{25 s}$ .

Solution

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

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, $v$ .

$\text{slope}={v}_{Q}=\frac{{\Delta x}_{Q}}{{\Delta t}_{Q}}=\frac{\left(\text{3120 m}-\text{1300 m}\right)}{\left(\text{32 s}-\text{19 s}\right)}$

Thus,

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

Discussion

This is the value given in this figure's table for $v$ at $t=\text{25 s}$ . The value of 140 m/s for ${v}_{Q}$ is plotted in [link] . 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 $a$ .

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

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

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try to read several books on phy don't just rely one. some authors explain better than other.
Ju
And don't forget to check out YouTube videos on the subject. Videos offer a different visual way to learn easier.
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