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The acceleration time graph of the motion of a particle from A to B is semicircle with radius of `sqrt(7)` . Find the velocity of particle at `t_(2)` at it `t_(1)` velocity is `3m//sec [pi = 22//7]` . A. `(15)/(2)m//sec`B. `8m//sec`C. `(17)/(2)m//sec`D. None of these |
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Answer» Correct Answer - C The acceleration ………….. we known that change in velocity = Area under a - t curve. `:. V_(t_(2)) - V_(t_(1))` = Area under the graph and t - axis from D to B (which is quaether circle) `:. V_(t_(2)) - V_(t_(1)) = (pi r^(2))/(4) = (22)/(7)xx((sqrt(7))^(2))/(4) =` `:. V_(t_(2))- 3=(11)/(2) or = (11)/(2) +3 = (17)/(2)m//sec` |
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