InterviewSolution
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InterviewSolution
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Acceleration of a particle in x-y plane varies with time as `a=(2t hati+3t^2 hatj) m//s^2` At time `t =0,` velocity of particle is `2 m//s` along positive x direction and particle starts from origin. Find velocity and coordinates of particle at `t=1s.`A. `(2t^(2) + 3t^(3)) W`B. `(2t^(2) + 4t^(4)) W`C. `(2t^(3) + 3t^(4)) W`D. `(2t^(3) + 3t^(5)) W` |
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Answer» `vecF = 2t hati + 3t^(2) hatj` `m (d vecv)/(dt) = 2t hati + 3t^(2) hatj (m = 1 kg)` `rArr underset(0) overset(vecv)intdvecv=underset(0)overset(t)int2thati+3t^(2) hat j " "(m=1kg)` Power `= vecF. Vecv = (2t^(3) + 3t^(5)) W` |
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