InterviewSolution
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InterviewSolution
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(a) If the initial velocity of a particle is u and collinear acceleration at any time `t` is at, calculate the velocity of the particle after time `t`. (b) A particle moves along a straight line such that its displacement at any time t is given by `s = (t^(3) - 6 t^(2) + 3t + 4)m`. What is the velocity of the particle when its acceleration is zero? |
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Answer» By definition acceleration `= (dv//dt)` So, `(dv)/(dt) = at` (given) or `int_(u)^(v) dv = int_(0)^(t) at` or `v - u = (1)/(2) a t^(2)` or `v = u + (1)/(2) a t^(2)` (b) As according to given problem, `s = t^(3) - 6t^(2) + 3t + 4` instantaneous velocity `u (ds)/(dt) = 3t^(2) - 12 t + 3` and acceleration `a = (dv)/(dt) = (d^(2)s)/(dt^(2)) = 6 t - 12`....(i) So, acceleration will be zero when `6t - 12 = 0`, i.e., `t = 2` sec And so the velocity when acceleration is zero, i.e., at `t = 2` sec from Eqn. (i), will be `v = 3 xx 2^(2) - 12 xx 2 + 3 = - 9m//s` [Negative velocity means that body is moving towards the origin, i.e., as time increases displacement decreases.] |
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