A particle moves along a staight line such that its displacement at any time t is given by `s=t^3-6t^2+3t+4m`. Find the velocity when the acceleration is 0.A. `3 ms^-1`B. `-12 ms^-1`C. `42 ms^-1`D. `-9 ms^-1`

A particle moves along a staight line such that its displacement at an

1.	A particle moves along a staight line such that its displacement at any time t is given by `s=t^3-6t^2+3t+4m`. Find the velocity when the acceleration is 0.A. `3 ms^-1`B. `-12 ms^-1`C. `42 ms^-1`D. `-9 ms^-1`
Answer» Correct Answer - D `s = t^2 - 6t + 3t + 4` Velocity `v = (ds)/(dt) = 3t^2 - 6 xx 2t + 3 xx 1 + 0` =`3t^2 - 12t + 3` ….(i) Acceleration `a = (dv)/(dt) = 3 xx 2t - 12 xx 1 + 0 = 6t - 12` Acceleration is zero at time t given by `6t-12=0` `rArr t = 2 seconds` So, velocity v at `t = 2` seconds is `v = (3t^2 - 12t + 3)_(t = 2s)` =`3 xx (2)^2 - 12 xx 2 + 3 = -9 m//s`.

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A particle moves along a staight line such that its displacement at any time t is given by `s=t^3-6t^2+3t+4m`. Find the velocity when the acceleration is 0.A. `3 ms^-1`B. `-12 ms^-1`C. `42 ms^-1`D. `-9 ms^-1`

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