A table clock has its minute hand ` 5.0 cm` long. Find the average velocity of the tip of the miute hand (a) between ` 6.000 a.m. ` to 6. 15 a.m. and (b) between ` 6.00 a.m. to 6.30. p.m.`

Explanation

(a) AT ` 6. 00 a.m.` the tip of the miute hand is at ` 12` mark and at ` 6. 15 a.m` it is ` 19^@` away. Thus, displacement
`= sqrt(R^2 + R^2 ) = sqrt 2 R = sqrt xx 5 cm`
Time taken from ` 6.00 a.m.` to ` 6.15 a.m. ` is `15 minutes = 15 xx 60 s`. The average velocity
` = ( displacement)/( tiem taken ) = ( sqrt 2 xx 5 cm)/( 15 xx 60 s)`
= 7. 86 xx 10^# cm s^-1`
Direction of average velocity is ` 45^@` with vertical. (b) AT ` 6.00 a.m.` the minute hand is at ` 12 ` mard and at ` 6. 30 p.m is ` 180^@` away. Thus, displacement ` =R+ R = 2 R = 2 xx 5 = 10 cm`. Time taken from ` 6.00 a.m. to ` 6. 30 p.m. is `12 ` hours and ` 30 minutes ` 45000 s`.
The averge velocity,
` v_(av0 = ( desplacement ) / ( tiem) = (10) /( 45 000)`
`= 2.2.xx 10^(-4) cm//s`
Direction of average velocity is `12 ` mark to the ` 6 ` mark on clock panel i.e. at ` 0^@` angle with vertical.