Question

In ${}^{\mathbf{1889}}$, at Jubbulpore, India, a tug-of-war was finally won after role="math" localid="1654754892705" ${}^{\mathbf{2}\mathbf{}\mathbf{h}\mathbf{}\mathbf{41}\mathbf{}\mathbf{min}}$, with the winning team displacing the center of the rope ${}^{\mathbf{3}\mathbf{.}\mathbf{7}\mathbf{}\mathbf{m}}$. In centimeters per minute, what was the magnitude of the average velocity of that center point during the contest?

Short answer

The magnitude of the average velocity of the center point of rope during the contest is${}^{\mathbf{2}\mathbf{.}\mathbf{3}\mathbf{}\mathbf{cm}\mathbf{/}\mathbf{min}}$

Step-by-step solution

Given data

Displacement, ${}^{\mathbf{\Delta x}\mathbf{}\mathbf{=}\mathbf{3}\mathbf{.}\mathbf{7}\mathbf{}\mathbf{m}}$

Time period, ${}^{\mathbf{\Delta t}\mathbf{}\mathbf{=}\mathbf{2}\mathbf{}\mathbf{h}\mathbf{}\mathbf{41}\mathbf{}\mathbf{min}}$

Understanding the average velocity

Converts the displacement into centimeters.

$$\begin{gathered} \mathbf{\Delta x}\mathbf{=}\mathbf{}\mathbf{3}\mathbf{.}\mathbf{7}\mathbf{\times }\mathbf{100}\mathbf{}\mathbf{cm} \\ \mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{=}\mathbf{370}\mathbf{}\mathbf{cm} \end{gathered}$$

Converts the time interval into minutes.

$$\begin{gathered} \mathbf{\Delta t}\mathbf{=}\left(2\times 60\right)\mathbf{}\mathbf{min}\mathbf{}\mathbf{+}\mathbf{41}\mathbf{}\mathbf{min} \\ \mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{=}\mathbf{120}\mathbf{}\mathbf{min}\mathbf{}\mathbf{+}\mathbf{41}\mathbf{}\mathbf{min} \\ \mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{=}\mathbf{161}\mathbf{}\mathbf{min} \end{gathered}$$

Using equation (i), the magnitude of average velocity of center point is calculated as follows:

$$\begin{gathered} {\mathbf{v}}_{\mathbf{avg}}\mathbf{=}\frac{\mathbf{\Delta x}}{\mathbf{\Delta t}} \\ \mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{=}\frac{\mathbf{370}\mathbf{}\mathbf{cm}}{\mathbf{161}\mathbf{}\mathbf{min}} \\ \mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{\approx }\mathbf{2}\mathbf{.}\mathbf{30}\mathbf{}\mathbf{min}\mathbf{/}\mathbf{cm} \end{gathered}$$

Thus, the average velocity of the center point by considering the displacement over the total time is${}^{\mathbf{2}\mathbf{.}\mathbf{30}\mathbf{}\mathbf{min}\mathbf{/}\mathbf{cm}}$