Question

Starting from a pillar, you run 200 m east (the +x-direction) at an average speed of 5.0 m/s and then run 280 m west at an average speed of 4.0 m/s to a post. Calculate (a) your average speed from pillar to post and (b) youraverage velocity from pillar to post.

Short answer

(a) The average speed from pillar to postis $v_{avg}^{speed}=4.36m/s$

(b) The average velocity from pillar to postis$v_{avg}^{velocity}=-0.73m/s$

Step-by-step solution

identification of given data

Distance covered in the east direction is,$s_1=200m$

The average speed in the east direction is,$v_1=5m/s$

Distance covered in the west direction is,$s_2=280m$

The average speed in the west direction is, $v_2=4m/s$

Calculationof the total time taken between pillar to post

The expression for the average speed is,

$v_{avg}^{speed}=\frac{s}{t}m/s$…………………..(i)

Where, and are the distance and time, respectively.

Substituting the given data in the equation (i) for the east direction,

$\begin{array}{rcl}{t}_1& =& \frac{s_1}{v_1}\\ & =& \frac{200m}{5m/s}\\ & =& 40s\end{array}$

Therefore, time taken from post to pillar in the east direction is, $t_1=40s$

Substituting the given data in the equation (i) for the west direction,

$\begin{array}{rcl}{t}_2& =& \frac{s_2}{v_2}\\ & =& \frac{280m}{4m/s}\\ & =& 70s\end{array}$

Therefore, time taken from pillar to post in the west direction is, $t_2=70s$

Calculationof the average speed

Total time taken in traveling between post to the pillar is,

$t=t_1+t_2$

Substituting the determined values of $t_1$ and $t_2$ in the above expression,

$\begin{array}{rcl}t& =& 40\text{s}+70s\\ & =& 110s\end{array}$

The total distance covered between pillar to post is,

$\begin{array}{rcl}s& =& s_1+s_2\\ & =& 200m+280m\\ & =& 480m\end{array}$

Substituting these values in equation (i) for calculation of average speed gives,

$\begin{array}{rcl}{v}_{avg}^{speed}& =& \frac{480m}{110s}\\ & =& 4.36m/s\end{array}$

Calculation for average velocity

The total displacement from pillar to post isthe difference between the distances traveled

between east and west directions, respectively,

Therefore, total displacement can be given as,

$\begin{array}{rcl}d& =& s_1-s_2\\ & =& 200m-280m\\ & =& -80m\end{array}$

The expression for the average velocity can be given as,

$v_{avg}^{velocity}=\frac{d}{t} m/s$…………………..(ii)

Where, and are the total displacement and total travel time, respectively.

Substituting the values of and in equation (ii) gives,

$\begin{array}{rcl}{v}_{avg}^{velocity}& =& \frac{d}{t}\\ & =& \frac{-80m}{110s}\\ & =& -0.73m/s\end{array}$

Thus the average velocity is, $v_{avg}^{velocity}=-0.73m/s$

Here negative sign shows that velocity is towards the west.