Question

Figure 4-21 shows the path taken by a skunk foraging for trash food, from initial point i. The skunk took the same time Tto go from each labeled point to the next along its path. Rank points a, b, and caccording to the magnitude of the average velocity of the skunk to reach them from initial point, greatest first. initial point i, greatest first.

Short answer

Rank is c=a>b

Step-by-step solution

Given information

$T-\mathrm{Time}\mathrm{to}\mathrm{travel}\mathrm{between}\mathrm{each}\mathrm{labeled}\mathrm{point}.$

$1\mathrm{block}=1\mathrm{unit}\mathrm{of}\mathrm{displacement}.$

To understand the concept

The problem deals with the average velocity which is the change in position or displacement divided by the time intervals in which the displacement occurs. Average velocity can be calculated by using displacement and time. Displacement is calculated by measuring change in position of object.

Formulae:

Average velocity is given by,

$v_{avg}=\frac{D}{T}$

Here$v_{avg}$ is the average velocity D, is the distance travelled andT is the time taken.

To rank the points a b c according to the average velocity

Point i to a:

D=1 block

Time=1T

Hence

$v_{avg}=\frac{1}{T}......\left(1\right)$

Point i to b:

D=1 block

Time=2T

Hence,

$v=\frac{1}{2T}$…… (2)

Point i to c

D =3block

Time=3T

Hence,

$v_{avg}=\frac{3}{3T}$

$\Rightarrow v_{avg}=\frac{1}{T}$…… (3)

Therefore, from these three results we can say that, rank according to average velocity would be $V_c=V_a>V_b$.