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(D) The length of the second's hand is $r = 6\,cm = 60\,mm$. The time period of the second's hand is $T = 60\,s$.The speed of the end point is given b

Vedclass · Accepted Answer

$6.28$ and $8.88\,mm/s$

Vedclass · Answer

(D) The length of the second's hand is $r = 6\,cm = 60\,mm$. The time period of the second's hand is $T = 60\,s$.The speed of the end point is given by $v = r\omega = \frac{r 	imes 2\pi}{T} = \frac{60\,mm 	imes 2\pi}{60\,s} = 2\pi\,mm/s \approx 6.28\,mm/s$.At two perpendicular positions,the velocity vectors $\vec{v_1}$ and $\vec{v_2}$ are perpendicular to each other,with $|\vec{v_1}| = |\vec{v_2}| = v = 6.28\,mm/s$.The magnitude of the difference in velocities is $|\Delta\vec{v}| = |\vec{v_2} - \vec{v_1}| = \sqrt{v_1^2 + v_2^2 - 2v_1v_2\cos(90^\circ)} = \sqrt{v^2 + v^2} = v\sqrt{2}$.$|\Delta\vec{v}| = 6.28 	imes 1.414 \approx 8.88\,mm/s$.

The second's hand of a watch has a length of $6\,cm$. The speed of the end point and the magnitude of the difference of velocities at two perpendicular positions will be:

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