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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?

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?

Grade:12th Pass

4 Answers

SOURAV MISHRA
37 Points
10 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
62 Points
10 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
11 Points
5 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
askIITians Faculty 646 Points
3 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

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