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 sec....Whate will be the position of this particle when it achieves max speed along +x direction?

x = 9t^2 - t^3.so v = 18t - 3t^2.for max. speed dv/dt = 0. this gives t = 3s.substituting in the expression for speed we get v =  27m/s at t  = 3s.so the position of the particle at t = 3s will be 54m. 

hi sonal

the velocity we will get by differentiating x in terms of t we get
v = 18 t - 3 t2        the RHS is quadratic expression of form a x2 + b x + c
will attain max value  when x = - b / 2a
a = -3      b = 18      c = 0     put this value  we will get
t = -18 / 2 (- 3) = 3
and put this value of t in the equation
x=9t2-t3
we will get x = 54

Find Distance travelled in first 10s Ans is 316m How to solve it and get ans 316m So please solve with detailed solution

Dear Student,
Please find below the solution to your problem.

Given that,
x = 9t^2 − t^3 .....(I)
We know that,
v = dx/dt​
The maximum speed is
v = 18t − 3t^2
v maximum, at dv/dt ​= 0
dv/​dt = 18 − 6t
The time will be
0 = 18−6t
t = 3sec
The position of the particle at 3 sec
Put the value of t in equation (I)
x = 9×9−27
x = 81−27
x = 54m
The position of the particle will be 54m.

Thanks and Regards