The second's hand of a watch has length $6\,\, cm$. Speed of end point and magnitude of difference of velocities at two perpendicular positions will be
The second's hand of a watch has length $6\,\, cm$. Speed of end point and magnitude of difference of velocities at two perpendicular positions will be
$6.28$ and $0 \,\,mm/s$
$8.88$ and $4.44 \,\,mm/s$
$8.88$ and $6.28 \,\,mm/s$
$6.28 $ and $8.88 \,\,mm/s$
The second's hand of a watch has length $6\,\, cm$. Speed of end point and magnitude of difference of velocities at two perpendicular positions will be
$v = r\omega = \frac{{r \times 2\pi }}{T} = \frac{{0.06 \times 2\pi }}{{60}} = 6.28\,mm/s$
Magnitude of change in velocity = $|\overrightarrow {{v_2}} - \overrightarrow {{v_1}} |$
$ = \sqrt {v_1^2 + v_2^2} = 8.88\,mm/s$ $\left( {{\rm{As\,}}{v_1} = {v_2} = 6.28\,mm/s} \right)$