Q115P - Fundamentals Of Physics

Question

In ${}^{\mathbf{1889}}$, at Jubbulpore, India, a tug-of-war was finally won after ${}^{\mathbf{2}\mathbf{ }\mathbf{h}\mathbf{ }\mathbf{41}\mathbf{ }\mathbf{min}}$, 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?

Step-by-Step Solution

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 1: Given data

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

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

Step 2: 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 ${}^{\mathbf{2}\mathbf{.}\mathbf{30}\mathbf{ }\mathbf{min}\mathbf{/}\mathbf{cm}}$