Grade 12th PassMechanics

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?

sonal sharma

12 Years agoGrade 12th Pass

4 Answers

SOURAV MISHRA

12 Years ago

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.

jitender lakhanpal

12 Years ago

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

Munaa

8 Years ago

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

Rishi Sharma

5 Years ago

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