Figure 2-18 shows four paths along which objects move from a starting point to a final point, all in the same time interval. The paths pass over a grid of equally spaced straight lines. Rank the paths according to (a) the average velocity of the objects and (b) the average speed of the objects, greatest

first.

The lines are equally spaced

The problem deals with the average speed and average velocity of an object. Average velocity is a vector quantity that has a direction and magnitude whereas the average speed is a scalar quantity.

Velocity is a vector quantity. According to the given graph, the lines are equally spaced which means the displacements of objects 1, 2, 3, and 4 are equal, all in the same time interval.

Therefore, objects 1, 2, 3, 4 have the same velocity.

So their ranks are,

$$\begin{gathered} \mathrm{Object}1\Rightarrow \mathrm{rank}1 \\ \mathrm{Object}2\Rightarrow \mathrm{rank}1 \\ \mathrm{Object}3\Rightarrow \mathrm{rank}1 \\ \mathrm{Object}4\Rightarrow \mathrm{rank}1 \end{gathered}$$

Since the time intervals are equal, speeds rank in order of total path length. Consider the average time is and the average displacement is .

Now, for object 1 the total distance traveled is

$1\mathrm{m}+1\mathrm{m}+1\mathrm{m}=3\mathrm{m}$

Thus the average speed of object 1 for 1 is 3 m/s

$1\mathrm{m}+1\mathrm{m}+1\mathrm{m}=3\mathrm{m}$

For object 3 the total distance traveled is

Thus the average speed of object 3 for 1 s is 1 m/s

For object 4 the total distance traveled is

$1\mathrm{m}+0.5\mathrm{m}+0.5\mathrm{m}+1\mathrm{m}+0.5\mathrm{m}+0.5\mathrm{m}+1\mathrm{m}=5\mathrm{m}$

Thus the average speed of object 4 for 1 s is 5 m/s

So, the rank will be

$\mathrm{path}4>\mathrm{path}1=\mathrm{path}2>\mathrm{path}3$