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The length of a second hand in watch is 1cm . The change in the velocity of tip in 15 sec...

The length of a second hand in watch is 1cm . The change in the velocity of tip in 15 sec...

Grade:12th pass

2 Answers

Arun
25750 Points
4 years ago
The tip of the second's hand moves (2×π×1)cm distance in 60 seconds. 
Thus, it's speed is (π/30)cm/s.
Let the initial position of the hand point at '12' in the clock. After 15 seconds, it'll point at '3'.
Let the initial velocity be V1, and final velocity be V2. The magnitude of change in velocity is |V2 - V1|. [The direction depends on the initial position of the second's hand, which hasn't been specified.]
Since both V1 and V2 have equal magnitude, it can be easily shown that the magnitude of the resultant vector Vris |-V1|*sqrt(2) or |V
2|*sqrt(2).
Here, |V1| = |V2| = (π/30)cm/s.
Thus the change in the magnitude of the velocity of the second's hand in 15 seconds is -
[{π*sqrt(2)}/30]cm/s
or, π/30 cm per sec is the answer
 
 
aswanth nayak
100 Points
4 years ago
Dear Student,
 
 
 
 
Change in velocity = change in distance/change in time
 
v= 1/100*15=0.0006666
 
hope this helps you
 
 
 
 
 
 
 
 
Regards
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 

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