Question

The length of a minute hand of a clock is 4 cm. Find the displacement and average velocity of the tip of the minute hand when it moves during a time interval from
(a) 3:15 pm to 3:30 pm (b) 4:15 pm to 4:45 pm.
A. (a) $\dfrac{{\sqrt 2 }}{{225}}\;{\rm{cm}}{{\rm{s}}^{ - 1}}$ (b) $\dfrac{2}{{225}}\;{\rm{cm}}{{\rm{s}}^{ - 1}}$
B. (a) $\dfrac{{\sqrt 3 }}{{225}}\;{\rm{cm}}{{\rm{s}}^{ - 1}}$ (b) $\dfrac{1}{{225}}\;{\rm{cm}}{{\rm{s}}^{ - 1}}$
C.(a) $\dfrac{{\sqrt 2 }}{{225}}\;{\rm{cm}}{{\rm{s}}^{ - 1}}$ (b) $\dfrac{1}{{225}}\;{\rm{cm}}{{\rm{s}}^{ - 1}}$
D. (a) $\dfrac{{\sqrt {12} }}{{225}}\;{\rm{cm}}{{\rm{s}}^{ - 1}}$ (b) $\dfrac{2}{{225}}\;{\rm{cm}}{{\rm{s}}^{ - 1}}$

Answer 1

Hint: The expression of the average velocity of the tip can be determined by dividing the displacement occurs in the position of the minute hand of the clock between its initial and final position with the total time taken by the minute hand of the clock.

Complete step by step solution:
Given:
The length of the minute hand of a clock is $l = 4\;{\rm{cm}}$.

(a)
The expression of the displacement of the minute hand is,
$D = \sqrt {{l^2} + {l^2}} $

Here $D$ is the displacement of the minute hand.

Substitute the value in the above expression
$\begin{array}{l}
D = \sqrt {4\;{\rm{cm}} + \;4\;{\rm{cm}}} \\
D = \sqrt 8 \;{\rm{cm}}\\
D = 4\sqrt 2 \;{\rm{cm}}
\end{array}$

The expression of the average velocity of the minute hand is,
${v_{avg}} = \dfrac{d}{t}$
Here $d$ is the displacement and $t$ is the total time.

Substitute the values in the above expression
$\begin{array}{l}
{v_{avg}} = \dfrac{{4\sqrt 2 \;{\rm{cm}}}}{{15\;{\rm{min}}\; \times \dfrac{{60\;{\rm{s}}}}{{\;{\rm{1 min}}}}}}\\
{v_{avg}} = \dfrac{{4\sqrt 2 \;{\rm{cm}}}}{{900\;{\rm{s}}}}\\
{v_{avg}} = {v_{avg}} = \dfrac{{\sqrt 2 \;}}{{225\;}}{\rm{cm}}{{\rm{s}}^{ - 1}}
\end{array}$

(b)
The expression of the displacement of the minute hand is,
$D = 2l$

Substitute the value in the above expression
$\begin{array}{l}
D = 2\left( {4\;{\rm{cm}}} \right)\\
D = 8\;{\rm{cm}}
\end{array}$

The expression of the average velocity of the minute hand is,
${v_{avg}} = \dfrac{d}{t}$

Substitute the values in the above expression
$\begin{array}{l}
{v_{avg}} = \dfrac{{8\;{\rm{cm}}}}{{30\;{\rm{min}}\; \times \dfrac{{60\;{\rm{s}}}}{{\;{\rm{1 min}}}}}}\\
{v_{avg}} = \dfrac{{8\;{\rm{cm}}}}{{1800\;{\rm{s}}}}\\
{v_{avg}} = \dfrac{1}{{225}}\;{\rm{cm}}{{\rm{s}}^{ - 1}}
\end{array}$

Therefore, the option (C) is the correct answer that is (a) $\dfrac{{\sqrt 2 }}{{225}}\;{\rm{cm}}{{\rm{s}}^{ - 1}}$ (b) $\dfrac{1}{{225}}\;{\rm{cm}}{{\rm{s}}^{ - 1}}$.

Note: For the calculation of the displacement in part (a), use the concept of Pythagoras and in part (b), minute hand makes 180 degree between its initial and final position, so for the displacement calculation just double the length of minute hand.