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Q.

The position $x$ of a particle with respect to time $t$ along x-axis is given by $x = 9 t^{2} - t^{3}$ 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?

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a

81m

b

32m

c

54m

d

24m

answer is A.

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Detailed Solution

$\begin{array}{l} v = \frac{d x}{d t} = 18 t - 3 t^{2} \\ a = \frac{d v}{d t} = 18 - 6 t \end{array}$

For maximum speed its acceleration should be zero 18-6t=0, t=3

Required position is x=81-27=54m

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The position x of a particle with respect to time t along x-axis is given by x=9t2−t3where 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?