A particle moves such that S = 6 + 48t - t^3. | vedclass

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(C) Given the position function $S = 6 + 48t - t^3$.The velocity $v$ is the rate of change of position with respect to time $t$,given by $v = \frac{dS

Vedclass · Accepted Answer

$134$

Vedclass · Answer

(C) Given the position function $S = 6 + 48t - t^3$.The velocity $v$ is the rate of change of position with respect to time $t$,given by $v = \frac{dS}{dt}$.$v = \frac{d}{dt}(6 + 48t - t^3) = 48 - 3t^2$.The direction of motion reverses when the velocity $v$ becomes zero.Setting $v = 0$,we get $48 - 3t^2 = 0$.$3t^2 = 48 \Rightarrow t^2 = 16 \Rightarrow t = 4$ (since time $t$ cannot be negative).To find the distance moved when the direction reverses,substitute $t = 4$ into the position equation $S$:$S(4) = 6 + 48(4) - (4)^3$.$S(4) = 6 + 192 - 64 = 134$.Thus,the particle reverses its direction after moving a distance of $134$ units.

$A$ particle moves such that $S = 6 + 48t - t^3$. The direction of motion reverses after moving a distance of

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