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A particle starts from point `A`, moves along a straight line path with an accleration given by `a = 2 (4 - x)` where `x` is distance from point `A`. The particle stops at point `B` for a moment. Find the distance `AB` (in `m`). (All values are in `S.I.` units) |
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Answer» Correct Answer - 8 `v(dv)/(dx) = 8 - 2x` `rArr overset(v)underset(0)(int)vdv = overset(x)underset(0)int(8 - 2x)dx rArr (v^(2))/(2) = 8x - x^(2)` `rArr v^(2) = 16x - 2x^(2)` At `B, v = 0` so, `x = 8` Hence, `AB = 8` |
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