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In the above example, the average acceleration of the particle in the interval t = 1 to t = 3 sec will be- |
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Answer» 12 a – 2b ` (dx)/(dt) = 2at - 3bt^(2)` Now velocity at t = 1 SEC, ` v_(1)= ((dx)/(dt))_(t=1) = 2a- 3B` and that at ` t = 3 sec , v_(2) = ((dx)/(dt))_(t= 3) = 6a - 27 b ` THUS average acceleration` a_(av) = (v_(2) - v_(1))/(t_(2) - t_(1))` ` = (6a - 27b - 2a + 3b)/(3-1) = (4a - 24b)/(2) = 2a - 12b ` Hence correct answer is (C) |
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