Question

The length of second's hand in a watch is $1cm$. Find the magnitude of change in velocity of its tip in $15$ seconds. Also find out the magnitude of average acceleration during this interval.

Answer 1

Radius, $r=1$ $cm={10}^{-2}$ $m$

Angular velocity, $\omega =\frac{2\pi }{T}$

$⟹\omega =\frac{2\pi }{60}$

$⟹\omega =0.105$ $rad/s$

$\therefore$ Linear velocity, $V=r\omega$

$⟹V={10}^{-2}\times 0.105$

$⟹V=0.00105$ $m/s$

$\therefore$ Change in velocity $=0.00105$ $m/s$

$\therefore$ Acceleration, $a=\frac{\Delta V}{\Delta t}$

$⟹a=\frac{0.00105}{15}$

$⟹a=0.00007$ $m/s^2$