Question

The length of a minute hand of a clock is 4 cm. Find the displacement and the average velocity of the tip of the minute hand when it moves from 3.15 p.m. to 3.30 p.m.

• A. $4\sqrt{2}cm,\frac{\sqrt{2}}{225}cm{s}^{-1}$
• B. $4\sqrt{2}cm,\frac{4}{225}cm$
• C. $4\sqrt{2}cm,\frac{1}{225}cm{s}^{-1}$
• D. $4\sqrt{2}cm,\frac{2}{225}cm{s}^{-1}$

Answer 1

The correct option is A $4\sqrt{2}cm,\frac{\sqrt{2}}{225}cm{s}^{-1}$

The displacement is shortest distance between the points

The the tip of the minute hand when it moves from 3.15 p.m. to 3.30 p.m makes a right angle triangle hence displacement is given by paythgoras theoram.

$d=\sqrt{4^2+4^2}=4\sqrt{2}cm$

The velocity is given by $v=\frac{displacement}{time}=\frac{4\sqrt{2}}{15min}=$$\frac{4\sqrt{2}}{15\ast 60}=\frac{\sqrt{2}}{225}cm/s$