A table clock has its minute hand 5.0 cm long.
Find the average velocity of the tip of the minute
hand (a) between 6.00 a.m. to 6.15 a.m. and
(b) between 6.00 a.m. to 6.30. p.m.
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Given :
Minute hand length =5cm
Average velocity =?
Time =6.00 am to 6.15 am
When the clock shows, 6.00 am , then the minute hand will point towards 12 and at 6.15 am then it shows at 90 degrees away from the before position.
Distance between initial and final position of tip of needle will be equal to displacement .
Displacement =hypotenuse of right angle triangle.
Displacement =√5² +5²
=7.07cm=7.07/100=0.071m
Total time= 15 minutes [from 6.am to 6.15am]
=15x60=900s
Average velocity= Total Displacement/total time
=0.071/900=7.88x10⁻⁵ m/s
∴ Average velocity of tip of the minute hand is 7.88x10⁻⁵ m/s
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