Question

A table clock has its minute hand $4cm$ long. Find the average velocity of the tip of the minute hand (a) between 6 AM to 6:30 AM and (b) between 6 AM to 6:30 PM

• A. $4\times {10}^{-3}\frac{cm}{s},1.8\times {10}^{-4}\frac{cm}{s}$
• B. $4.4\times {10}^{-3}\frac{cm}{s},1.8\times {10}^{-4}\frac{cm}{s}$
• C. $0.44\times {10}^{-3}\frac{cm}{s},1.8\times {10}^{-4}\frac{cm}{s}$
• D. $4.4\times {10}^{-3}\frac{cm}{s},0.18\times {10}^{-4}\frac{cm}{s}$

Answer 1

The correct option is B $4.4\times {10}^{-3}\frac{cm}{s},1.8\times {10}^{-4}\frac{cm}{s}$

At 6 AM the tip of the minute hand is at 12 mark and at 6:30 AM or 6:30 PM it is ${180}^{\circ }$ away. Thus, the straight-line distance between the initial and final positions of the tip of minute hand is equal to the diameter of the clock.

Displacement $=2\times R=2\times 4cm=8cm$
The displacement is from 12 mark to 6 mark on the clock panel. This is also the direction of the average velocity in both the cases. Therefore,

(a) The time taken from 6 AM to 6:30 AM is $30min=1800s$. The average velocity is
$v_{av}=\frac{Displacement}{Time}=\frac{8}{1800}=4.4\times {10}^{-3}\frac{cm}{s}$

(b) The time taken from 6 AM to 6:30 PM is $12.5hr=45000s$. The average velocity is
$v_{av}=\frac{Displacement}{Time}=\frac{8}{45000}=1.8\times {10}^{-4}\frac{cm}{s}$