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(C) The distance covered by the particle is given by $s = 2 + 27t - t^3$.To find when the particle stops,we need to find the time $t$ when its velocit

Vedclass · Accepted Answer

$56 	ext{ meters}$

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(C) The distance covered by the particle is given by $s = 2 + 27t - t^3$.To find when the particle stops,we need to find the time $t$ when its velocity $v = \frac{ds}{dt}$ is zero.$\frac{ds}{dt} = \frac{d}{dt}(2 + 27t - t^3) = 27 - 3t^2$.Setting the velocity to zero: $27 - 3t^2 = 0$.$3t^2 = 27 \Rightarrow t^2 = 9$.Since time $t > 0$,we have $t = 3 	ext{ seconds}$.Now,we calculate the distance covered at $t = 3 	ext{ seconds}$:$s(3) = 2 + 27(3) - (3)^3$.$s(3) = 2 + 81 - 27$.$s(3) = 56 	ext{ meters}$.Therefore,the particle will stop after covering $56 	ext{ meters}$.

The distance $s$ in meters covered by a particle in $t$ seconds is given by $s = 2 + 27t - t^3$. The particle will stop after covering a distance of:

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