The position of a particle moving along the x axis depends on the time according to equation x = 3t²-2t³

At what time does the particle reach its maximum positive x position? And when does it reach minimum?

(I believe this requires calculus. Could someone show me?)

Hi Pritish,

Remember that at the point of maxima and minima, dy/dx = 0

and if d²y/dx² is -ve, then that value of x gives maxima and if,

d²y/dx² = +ve, then it is minima.

Given, x = 3t² - 2t³

dx/dt = 6t - 6t²

For maxima, or minima,

dx/dt = 0

6t - 6t² = 0

t = t²

t (t - 1) = 0

t = 0s & t = 1 sec

Let us find d²x/dt²

(d²x/dt² )_{t=0  =}  6 - 12t      

                       =   6 - 12 x 0

                       =  6+ve

(d²x/dt² )_{t=1  =  6 - 12 x 1}                               

                       = -6

                       =  - ve

Hence, at t = 1 sec, particle reaches its maximum +ve  x position and at x = 0 it reaches minimum x - position