The equation of motion of a particle moving on circular path (radius 200m) is given by $s = 18 t + 3 t^{2} - 2 t^{3}$ where $s$ is the total distance covered from straight point in meters at the end of $t$ seconds. The maximum speed of the particle will be:

15 m / s e c

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23 m / s e c

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19.5 m / s e c

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25 m / s e c

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Solution

The correct option is B $19.5 m / s e c$
The equation of motion of a particle on the circular path is given by
$s = 18 t + 3 t^{2} - 2 t^{3}$
Speed, $v = \frac{d s}{d t} = 18 + 6 t - 6 t^{2}$ . . . . .(1)
For maximum speed, $\frac{d v}{d t} = 0$
$6 - 12 t^{=} 0$
$t = \frac{1}{2} = 0.5 s$
The maximum speed of the particle will be
$v_{m a x} = 18 + 6 \times 0.5 - 6 \times 0.5 \times 0.5$
$v_{m a x} = 18 + 3 - 1.5$
$v_{m a x} = 19.5 m / s$
The correct option is C.

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