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Chapter 2: Q115P (page 39)

In 1889, at Jubbulpore, India, a tug-of-war was finally won after role="math" localid="1654754892705" 2h41min, with the winning team displacing the center of the rope 3.7m. In centimeters per minute, what was the magnitude of the average velocity of that center point during the contest?

Short Answer

Expert verified

The magnitude of the average velocity of the center point of rope during the contest is2.3cm/min

Step by step solution

01

Given data

Displacement, Δx=3.7m

Time period, Δt=2h41min

02

Understanding the average velocity

Converts the displacement into centimeters.

Δx=3.7×100cm=370cm

Converts the time interval into minutes.

Δt=(2×60)min+41min=120min+41min=161min

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

vavg=ΔxΔt=370cm161min≈2.30min/cm

Thus, the average velocity of the center point by considering the displacement over the total time is2.30min/cm

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