A force acts on a 3 g particle in such a way that the position of the particle as a function of time

Question

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 $x=3 t-4 t^{2}+t^{3}$, where $x$ is in metres and $t$ is in seconds. The work done during the first 4 second is\

Professor PC Thomas Classes · Answer

(d) : $x=3 t-4 t^{2}+t^{3}$\ $V=\frac{d x}{d t}=3-8 t+3 t^{2}$\ $V_{0}=3-8 	imes 0+3 	imes 0^{2}=3 \mathrm{~m} / \mathrm{s}$\ $V_{4}=3-8 	imes 4+3 	imes 16=19 \mathrm{~m} / \mathrm{s}$\ $W=\frac{1}{2} m\left(V_{4}^{2}-V_{0}^{2}ight)=\frac{1}{2} 	imes 3 	imes 10^{-3}\left(19^{2}-3^{2}ight)$ $$ =528 	imes 10^{-3} \mathrm{~J} \approx 530 \mathrm{~mJ} $$