Larry leaves home at 9:05 and runs at a constant speed to the lamp post seen in FIGURE EX2.3. He reaches the lamppost at 9:07, immediately turns, and runs to the tree. Larry arrives at the tree at 9:10. a. What is Larry’s average velocity, in m/min, during each of these two intervals? b. What is Larry’s average velocity for the entire run?

Time taken by Person L to travel between house to a lamp post is $t_1=9:07-9:05=2min$

Displacement from house to a lamp post is $x_1=600-200=400m$. Here displacement towards the left is considered positive.

Time taken by Person L to travel from lamp post to tree is $t_2=9:10-9:07=3min$

Displacement from the lamp post to the tree is localid="1648452223669" $x_2=-(1200-200)=-1000m$

Average velocity is given by

Average velocity $=\frac{Totaldisplacement}{totaltimetaken}$

Consider the first interval, that is, from house to lamp post

The average velocity of person L is

$$\begin{gathered} v_1=\frac{x_1}{t_1} \\ =\frac{400}{2} \\ =200m/min \end{gathered}$$

Therefore, the average velocity of Person L in the first interval is $200m/min$.

Consider a second interval, that is, from lamp post to the tree

The average velocity of Person L is

localid="1648452295972" $$\begin{gathered} v_2=\frac{x_2}{t_2} \\ =\frac{-1000}{3} \\ =-333.3m/min \end{gathered}$$

Therefore, the average velocity of Person L in the second interval islocalid="1648452322743" role="math" $-333.3m/min$.

The total displacement traveled is

$$\begin{gathered} x=x_1+x_2 \\ =400-1000 \\ =-600m \end{gathered}$$

The total time taken is

$$\begin{gathered} t=t_1+t_2 \\ =2+3 \\ =5min \end{gathered}$$

The average velocity of Person L for the entire run is

localid="1648452438534" $$\begin{gathered} v=\frac{x}{t} \\ =\frac{-600}{5} \\ =-120m/min \end{gathered}$$

Therefore, the average velocity of the person L for the entire run is localid="1648452465548" $-120m/min$.