Figure 2-20 gives the velocity of a particle moving along an axis. Point 1 is at the highest point on the curve; point 4 is at the lowest point; and points 2 and 6 are at the same height. What is the direction of travel at (a) time $\mathbf{t}\mathbf{=}\mathbf{0}$and (b) point 4? (c) At which of the six numbered points does the particle reverse its direction of travel? (d) Rank the six points according to the magnitude of the acceleration, greatest first.

The graph in figure 2.20

The velocity is the rate of change of displacement. The positive value of the velocity implies that the change in the displacement is positive. It means the particle is traveling in a positive direction. Similarly, the negative value of velocity implies that the particle is traveling in the negative direction.

The velocity of the particle is given by,

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

Here $d\overrightarrow{x}$is the change in the displacement.

From the graph, it can be seen that the value of the velocity at t = 0 is positive. Thus, the direction of travel at the time t = 0 is in the positive direction of the axis.

From the graph, it can be seen that the velocity at point 4 is negative. Thus, the direction of travel at point 4 is in the negative direction of the axis

From the graph it is seen that the velocity changes from positive to negative at point 3 and negative to positive at point 5. Thus, the particle reverses its direction of travel at points 3 and 5

Here the magnitude of acceleration is given by the slope of the graph. From the graph, the slope of the graph at points 1 and 4 is zero.

The slope of the graph is negative at points 2 and 3 with the slope being more negative at 2 than 3.

The slope of the graph is positive at points 5 and 6 with the slope being more positive at 6 than 5.

Therefore, the ranking is$6>5>4=1>3>2$