A particle is moving in a straight line. At time $t$ , the distance between the particle from its starting point is given by $x = t - 6 t^{2} + t^{3}$ . Its acceleration will be zero at

$t = 1 u n i t t i m e$

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$t = 2 u n i t s t i m e$

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$t = 3 u n i t s t i m e$

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$t = 4 u n i t s t i m e$

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Solution

The correct option is B
$t = 2 u n i t s t i m e$

Step 1: Given data
Distance, $x = t - 6 t^{2} + t^{3}$
Step 2: Find time at which acceleration will be zero
Acceleration, $a$ is equal to the second derivative of the distance or the displacement.
$x = t - 6 t^{2} + t^{3}$
Differentiating both the sides, we get
$\frac{d x}{d t} = 1 - 12 t + 3 t^{2}$
Again, differentiating both the sides, we get
$a = \frac{d^{2} x}{d t^{2}} = - 12 + 6 t$
As acceleration is zero,
$a = 0 - 12 + 6 t = 0 t = 2 u n i t s t i m e$
Hence, option $(B)$ is correct.

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A particle is moving in a straight line. At time t, the distance between the particle from its starting point is given by x=t-6t2+t3. Its acceleration will be zero at

A particle is moving in a straight line. At time $t$ , the distance between the particle from its starting point is given by $x = t - 6 t^{2} + t^{3}$ . Its acceleration will be zero at