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The position $x$ of a particle with respect to time $t$ along x-axis is given by $x = 9t^2 - t^3$ where $x$ is in metres and $t$ in second. What will be the position of this particle when it achieves maximum speed along the +ve x direction?
A
54 m
B
81 m
C
24 m
D
32 m
Detailed Solution
Speed: $v = \dfrac{dx}{dt} = \dfrac{d}{dt}(9t^2 - t^3) = 18t - 3t^2$
For maximum speed, $\dfrac{dv}{dt} = 0$
$\dfrac{dv}{dt} = 18 - 6t = 0 \Rightarrow t = 3$ s
Since $\dfrac{d^2v}{dt^2} = -6$ is negative, this is a maximum.
Position at $t = 3$ s: $x = 9(3)^2 - (3)^3$
$x = 81 - 27 = 54$ m
So the particle is at 54 m when it achieves its maximum speed along the +ve x direction.
For maximum speed, $\dfrac{dv}{dt} = 0$
$\dfrac{dv}{dt} = 18 - 6t = 0 \Rightarrow t = 3$ s
Since $\dfrac{d^2v}{dt^2} = -6$ is negative, this is a maximum.
Position at $t = 3$ s: $x = 9(3)^2 - (3)^3$
$x = 81 - 27 = 54$ m
So the particle is at 54 m when it achieves its maximum speed along the +ve x direction.