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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?)

9 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

9 years ago
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