39D = 13.8 ± 0.2 m
T = 4 ± 0.3 s
S = D/T = 13.8/4 = 3.45
∂S/S = ±(∂D/D + ∂T/T)
=> ∂S/S = ±(0.2/13.8 + 0.3/4)
=> ∂S = ±0.308 ≈ ±0.31
So S = 3.45 ± 0.31 m/s
A BODY TRAVELS UNIFORMLY A DISTANCE OF (13.8±0.2) m IN A TIME OF (4.0±0.3) s. THE VELOCITY OF THE BODY IS.
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7 Answers
39
1MY DOUBT IS TIME HAS 2 SIGNIFICANT FIGURES , SHOULDN'T VELOCITY (S) HAVE 2 SIGNIFICANT FIGURES i.e. S=3.4±0.3
1THAT WAS WHAT GIVEN IN THE BOOK ,I JUST WANTED TO CLEAR MY DOUBT ,THANKS.
1ANOTHER ONE : IF ERROR IN MEASUREMENT OF MOMENTUM OF A PARTICLE IS +100% THEN ERROR IN MEASUREMENT OF KINETIC ENERGY IS.
29see this thread for the answer
http://www.targetiit.com/iit-jee-forum/posts/all-time-doubt-service-15095.html