Dear Student, a displacement of the minute hand tipd=8 cm because it reachec just oppsite to the diametric end time taken=30*60=1800 svav=81800=4 4*10-3 cm /sb the displacement remain same that is d=8 cm time taken =12 hr 30 min=45000 saverage velcity=845000=1 8*10-4 cm /sRegards

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Solve this: of a particle in the time interval written as (3.4) Note that the BBitions of the particle at time And t t are used in calculating the average The v»sitions in twtween tt and t: are not hence the actual path taken in going from A B not important in calculating the average A has its minute hand Long. Find the the tip of the minute hand (a) a-m to 6-30 a.m. and (b) 600 a-m. to : At the tip the minute hand is at 12 mark and at 6-30 a.m. or 630 p.m. it is ISO' away. the straight line distance between the initial and of the tip ts equal to the diameter of the 2 R = 2 x 4-0 = em- The displacement is from the 12 mark to the 6 mark on panel. This is the of the vekvit.v in tx•th ea—. The time taken from 600 a-m_ to a.m- 30 minutes = ISOO s- The average velæity is Displacement SO cm = 44 x - ISOOs Te time taken from 600 a-m- to 6-30 p.m. 12 and minutes 45000 s. The aver* vel.xity SO cm In be of In

Dear Student,

a . d i s p l a c e m e n t o f t h e m i n u t e h a n d t i p d = 8 cm because it reachec just oppsite to the diametric end . time taken = 30 * 60 = 1800 s v_{av} = \frac{8}{1800} = 4.4 * 10^{- 3} cm / s b . the displacement remain same that is d = 8 cm time taken = 12 hr 30 \min = 45000 s average velcity = \frac{8}{45000} = 1.8 * 10^{- 4} cm / s Regards

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