91Ó°ÊÓ

Distance walked from house to the windmill is,$s_1=60.0\text{m}$

Distance walked from the windmill to the bench is,$s_2=40.0\text{m}$

Time takenin walking from house to windmill is,$t_1=28.0\text{s}$

Time taken in walking from the windmill to the bench is,$t_2=36.0s$

(a)

Total displacement from the front door is,

$\begin{array}{rcl}d& =& s_1-s_2\\ & =& 60\text{m}-40\text{m}\\ & =& 20 \text{m}\\ & & \end{array}$

This 20 m is the shortest distance between your front door and the bench.

The total time taken by you to go to the windmill from the house and then back to the bench from the windmill is

$\begin{array}{rcl}t& =& t_1+t_2\\ & =& 28.0\text{s}+36.0\text{s}\\ & =& \text{64.0} \text{s}\\ & & \end{array}$

The expression for the average velocity can be given as,

$v_{avg}^{velocity}=\frac{\text{displacement}}{\text{time}}$

Substituting the values in the above expression,

$\begin{array}{rcl}{v}_{avg}^{velocity}& =& \frac{20m}{0.64.0s}\\ & =& 0.32m/s\end{array}$

Thus, your average velocity is $v_{avg}^{velocity}=0.32 m/s$

(b)

The total distance covered by you in walking from your house to the windmill and then to the bench is,

$\begin{array}{rcl} s& =& s_1+s_2\\ & =& 60\text{m}+40\text{m}\\ & =& 100 \text{m}\\ & & \end{array}$

The expression for the average velocity can be given as,

$v_{avg}^{speed}=\frac{\text{total distance}}{\text{time}}$

Substituting the values in the above expression,

$\begin{array}{rcl}{v}_{avg}^{speed}& =& \frac{100m}{64s}\\ & =& 1.56m/s\end{array}$

Thus, your average speed is $v_{avg}^{speed}=1.56m/s$