A force acts on a 3, g particle in such a way | vedclass

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(C) Given position: $x = 3t - 4t^2 + t^3$.Velocity is the derivative of position with respect to time: $v = \frac{dx}{dt} = 3 - 8t + 3t^2$.At $t = 0\,

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$528$

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(C) Given position: $x = 3t - 4t^2 + t^3$.Velocity is the derivative of position with respect to time: $v = \frac{dx}{dt} = 3 - 8t + 3t^2$.At $t = 0\, s$,the initial velocity $v_1 = 3 - 8(0) + 3(0)^2 = 3\, m/s$.At $t = 4\, s$,the final velocity $v_2 = 3 - 8(4) + 3(4)^2 = 3 - 32 + 48 = 19\, m/s$.According to the Work-Energy Theorem,the work done $W$ is equal to the change in kinetic energy: $W = \Delta K = \frac{1}{2}m(v_2^2 - v_1^2)$.Given mass $m = 3\, g = 3 	imes 10^{-3}\, kg$.$W = \frac{1}{2} 	imes (3 	imes 10^{-3}) 	imes (19^2 - 3^2)$.$W = \frac{1}{2} 	imes 3 	imes 10^{-3} 	imes (361 - 9) = \frac{1}{2} 	imes 3 	imes 10^{-3} 	imes 352$.$W = 3 	imes 10^{-3} 	imes 176 = 528 	imes 10^{-3}\, J = 528\, mJ$.

$A$ force acts on a $3\, g$ particle in such a way that the position of the particle as a function of time is given by $x = 3t - 4t^2 + t^3$,where $x$ is in $meters$ and $t$ is in $seconds$. The work done during the first $4\, seconds$ is .............. $mJ$.

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