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In 1889, at Jubbulpore, India, a tug-of-war was finally won after 2 h 41 min, with the winning team displacing the center of the rope 3.7 m. What was the magnitude of the average velocity of that center point during the contest

Answer :

Answer:

The magnitude of the average velocity of that center point during the contest is 0.0229 [tex]\frac{m}{min}[/tex]

Explanation:

Given :

Time interval[tex]\Delta t = 161[/tex] min

Displacement of the rope [tex]\Delta x = 3.7[/tex] m

For calculating average velocity of the rope,

  [tex]v = \frac{\Delta x}{\Delta t}[/tex]

  [tex]v = \frac{3.7}{161}[/tex]

  [tex]v = 0.0229[/tex] [tex]\frac{m}{min}[/tex]

Therefore, the magnitude of the average velocity of that center point during the contest is 0.0229 [tex]\frac{m}{min}[/tex]

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