Class 11 NEET Physics Motion in a Straight Line Questions

$x=40+12 t-t^{3}$ $\therefore$ Velocity, $v=\frac{dx}{dt}=12-3 t^{2}$ When particle come to rest, $\frac{dx}{dt}=v=0$ $\therefore 12-3 t^{2}=0 \Rightarrow 3 t^{2}=12$ $\Rightarrow t=2 \text{ sec}$ Distance travelled by the particle before coming to rest $\int\limits_{0}^{s} ds=\int\limits_{0}^{2} v dt$ or $s=\int\limits_{0}^{2} (12-3 t^{2} ) dt$ $=\left(12 t- \frac{3 t^{3}}{3} \right)\Big|_{0} ^{2}$ $s=12 \times 2-8=24-8$ $=16 \text{ m}$ (OR) At $t=0$, particle is at , let's say $x$ distance, from $O$ then putting $t=0$ in the given displacement-time equation we get; $x=40+12(0)-(0)^{3}=40 \text{ m}$ Particle comes to rest that means velocity of particle becomes zero after travelling certain displacement ; let's say the time be t. then after differentiating the given displacement-time equation w.r.t. time we get velocity - time equation $v=12-3 t^{2}$ when the particle comes to rest ): at time $t=t$ $v=0$ $=>12-3 t^{2}=0$ $=>t=2 s$ Then, at $t=2$ s we are at , let's say $x^{'}$ distance from $O$; put this value oft $(2)$ in given displacement-time equation, we get; $x^{'}=40+12(2)-(2)^{3}$ $=56 \text{ m}$ Further; We have seen that the particle started his journey when it is at $40 \text{ m}$ from the point $O$. And came to rest at $56 \text{ m}$ from the point $O$ then the particle traveled a distance of: $56-40=16 \text{ m}$