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(B) The displacement of the particle is given by $s = t^3 - 6t^2 + 3t + 4$. Velocity $v$ is the first derivative of displacement with respect to time:

Vedclass · Accepted Answer

$-9$

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(B) The displacement of the particle is given by $s = t^3 - 6t^2 + 3t + 4$. Velocity $v$ is the first derivative of displacement with respect to time: $v = \frac{ds}{dt} = 3t^2 - 12t + 3$. Acceleration $a$ is the derivative of velocity with respect to time: $a = \frac{dv}{dt} = 6t - 12$. To find the time when acceleration is zero,set $a = 0$: $6t - 12 = 0 \implies t = 2 \ s$. Now,substitute $t = 2 \ s$ into the velocity equation: $v = 3(2)^2 - 12(2) + 3 = 3(4) - 24 + 3 = 12 - 24 + 3 = -9 \ m/s$.

$A$ particle moves along a straight line such that its displacement at any time $t$ is given by $s = (t^3 - 6t^2 + 3t + 4) \ m$. Find the velocity of the particle in $m/s$ when its acceleration is zero.

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