First slide
Instantaneous velocity
Question

The position x of a particle with respect to time t along X-axis is given by x = 9t2 - t3, where x is in metres and t in second. What will he the position of this particle when it achieves maximum speed along the positive x-direction?

Moderate
Solution

The position x of a particle w.r.t. time t along X-axis,

                 x = 9t2 - t3                                                      …(i)

Differentiating Eq. (i) w.r.t. time, we get speed, i.e.

                v=dxdt=ddt9t2-t3

           v = 18t - 3t2                                                     …(ii)

Again, differentiating Eq. (ii) w.r.t. time, we get acceleration, i.e.

                 a=dvdt=ddt18t-3t2

            a = 18 - 6t                                                        …(iii)

Now, when speed of particle is maximum, its acceleration is zero.

                  a = 0

         18 - 6t = 0

                     t = 3 s

Putting in Eq. (i), we obtain position of particle at that time

                      x=9(3)2-(3)3=9(9)-27

                         = 81 - 27 = 54 m

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