Q5E - University Physics with Modern Physics

Question

Starting from the front door of a ranch house, you walk 60.0 m due east to a windmill, turn around, and then slowly walk 40.0 m west to a bench, where you sit and watch the sunrise. It takes you 28.0 s to walk from the house to the windmill and then 36.0 s to walk from the windmill to the bench. For the entire trip from the front door to the bench, what are your (a) average velocity and (b) average speed?

Step-by-Step Solution

Answer

(a) Your average velocity is,$v_{avg}^{velocity}=0.32 m/s$ 

(b) Your average speed is, $v_{avg}^{speed}=1.56 m/s$

Step 1: Identification of the given data

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, $t_2=36.0s$

Step 2: Calculation for the average velocity

(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  $v_{avg}^{velocity}=0.32 m/s$

 

Step 3: Calculation of the average speed

(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  $v_{avg}^{speed}=1.56m/s$