The equation of motion of a particle moving along a straight line is $s = 2 t^{3} - 9 t^{2} + 12 t$ , where the units of s and t are centimetre and second. The acceleration of the particle will be zero after.

\frac{3}{2} s

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\frac{2}{3} s

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\frac{1}{2} s

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1 s

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Solution

The correct option is A $\frac{3}{2} s$
Given, $s = 2 t^{3} - 9 t^{2} + 12 t$
$\frac{d s}{d t} = 6 t^{2} - 18 t + 12$

Again, $\frac{d^{2} s}{d t^{2}} = 12 t - 18 =$ acceleration.

If acceleration becomes zero, then $0 = 12 t - 18$
$\Rightarrow t = \frac{3}{2} s$

Hence, acceleration will be zero after $\frac{3}{2} s$ .

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