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Pritish Chakraborty Grade: 12
        

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?)

8 years ago

Answers : (1)

askIITIians Expert
21 Points
										

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

8 years ago
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