Question

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 1

The magnitude of the average velocity of that center point during the contest is 0.0229
`(m)/(min)`

Step-by-step explanation:

Given :

Time interval
`\Delta t = 161` min

Displacement of the rope
`\Delta x = 3.7` m

For calculating average velocity of the rope,

`v = (\Delta x)/(\Delta t)`

`v = (3.7)/(161)`

`v = 0.0229`
`(m)/(min)`

Therefore, the magnitude of the average velocity of that center point during the contest is 0.0229
`(m)/(min)`