Answer to Question #290522 in Physics for dede

Question

The position of a particle moving along x axis is given by x=6.0t²-1.0t² where x is in meter and t is in seconds. What is the position of the particle when it achieves its maximum speed is the positive x direction?

Answer 1

The position of a particle is given by

$$x=6.0t^2-1.0t^3$$

The speed of particle

$$v=(x)'=12t^2-3.0t$$

The acceleration

$$a=(v)'=24t-3.0$$

The particle achieves its maximum speed when $a=0$, so

$$24t-3.0=0,\quad t=8\:\rm s$$

The position of particle at this instant

$$x=6.0*8^2-1.0*8^3=-128\:\rm m$$