91Ó°ÊÓ

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,

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

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

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