Question

Length of a minute hand of a clock is $4.5cm$. Find average velocity of time of the minute hand between $6AM$. to $6.30A.M$ and $6A.M.$ to $6.30P.M$.

Answer 1

Given that,

Time $t=30min=1800sec$

Now, the length of the minute hand is almost as long as the radius of the clock

The displacement is

$D=2\times 4.5$

$D=9cm$

$D=0.09m$

Now, the velocity is

$v=\frac{D}{t}$

$v=\frac{0.09}{1800}$

$v=5\times {10}^{-5}m/s$

Hence, the average velocity is $5\times {10}^{-5}m/s$

Displacement during $6:00AM$ to $6:30PM$ will be:

$D=4.5\times 2=9cm$

$D=0.09m$

Time taken:

$t=12.5$ Hours.

$t=12.5\times 3600$ sec

Therefore,

Velocity will be given by:

$v=\frac{0.09}{12.5\times 3600}=2\times {10}^{-6}m/s$