Figure 2-19 is a graph of a particle’s position along an x axis versus time. (a) At time$\mathit{t}\mathbf{=}\mathbf{0}$, what is the sign of the particle’s position? Is the particle’s velocity positive, negative, or 0 at (b)$\mathit{t}\mathbf{=}\mathbf{1}\mathbf{}\mathit{s}$, (c)$\mathit{t}\mathbf{=}\mathbf{2}\mathbf{}\mathbf{s}$, and (d)$\mathit{t}\mathbf{=}\mathbf{3}\mathbf{}\mathbf{s}$? (e) How many times does the particle go through the point$\mathbf{x}\mathbf{=}\mathbf{0}$?

The graph in figure 2.19

The velocity is the rate of change of displacement. If the displacement for the given time interval is positive, the velocity is also positive, if the displacement is negative the velocity is negative. The velocity is also equal to the slope of displacement vs the time graph.

From the given graph it is seen that when t =0, the position of the particle is below the time axis. It means the position is on the negative axis of the position. Thus, the position of the particle at t =0 is negative

The velocity of the particle is given by,

$\overrightarrow{v}=\frac{d\overrightarrow{x}}{dt}$

The expression also represents the slope of the graph. Now, if the slope is positive, it can be concluded that the velocity is positive. If the slope is negative, it can be concluded that the velocity is negative. And, if the slope is zero, it can be concluded that the velocity is zero.

From the graph, it can be seen that the slope is positive. Thus, the velocity of the particle at t =1 is positive.

The slope of x versus t is zero.

Thus, the velocity of the particle att=2 s is zero

From the graph, it can be seen that the slope of x versus t is negative.

Thus, the velocity of the particle at t =3 s is negative.

From the graph, it can be seen that the particle goes through the pointx=0 twice at t =1 s and t =3 s