Question

4) The position x of a particle with respect to time t along the X-axis is given by x = 9t^2 - t^3 where
x is in metre and t in second. What will be the position of this particle when it achieves
maximum speed along the +X direction?​

Answer 1

Answer: 54 m

Explanation:

$x=9t^{2}-t^{3}$

or, v=$\frac{dx}{dt}=\frac{d(9t^{2}- t^{3})}{dt}$

or, v=18t-$3t^{2}$

v will be minimum or maximum when $\frac{dv}{dt}$ will be zero.

∴ $\frac{dv}{dt}$=0

or, $\frac{d(18t-3t^{2}) }{dt}$=0

or, 18-6t=0

or, 6t=18

or, t=18/6=3

at t=3 sec x=$9(3)^{2}-(3)^{3}$=9*9-27=81-27=54m.

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