If the distance ‘ $s$ ’ metre traversed by a particle in $t$ seconds is given by $s = t^{3} - 3 t^{2}$ , then the velocity of the particle when the acceleration is zero, in $metre / \sec$ is,

$3$

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$- 2$

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$- 3$

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$2$

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Solution

The correct option is C
$- 3$

Step 1: Given data
The distance ‘ $s$ ’ metre traversed by a particle in $t$ seconds is, $s = t^{3} - 3 t^{2}$
Step 2: Formula used
The first derivative of displacement is velocity
That is, $v = \frac{ds}{dt}$
Acceleration( $a$ ) is the derivative of velocity with respect to time.
That is, $a = \frac{dv}{dt}$
In other words, acceleration( $a$ ) is the second derivative of displacement( $s$ ).
That is, $a = \frac{d^{2} s}{{dt}^{2}}$
Step 3: Find Velocity ( $v$ ):
Velocity( $v$ ) is the rate of change of distance of an object with respect to time.
Here, $s = t^{3} - 3 t^{2}$
Velocity, $v = \frac{ds}{dt}$
That is,
$v = \frac{d (t^{3} - 3 t^{2})}{dt} v = 3 t^{2} - 3 \times 2 t v = 3 t^{2} - 6 t . . . . . (1)$
Step 4: Find Acceleration ( $a$ ):
Acceleration( $a$ ) is the rate of change of the velocity of an object with respect to time.
Acceleration, $a = \frac{dv}{dt}$
That is,
$a = \frac{d (3 t^{2} - 6 t)}{dt} a = 2 \times 3 t - 6 a = 6 t - 6 . . . . (2)$
Step 5: Find Velocity at $a = 0$ :
To find the time t where the acceleration becomes zero, equate the equation (2) to zero
$a = 6 t - 6 0 = 6 t - 6 6 = 6 t \frac{6}{6} = t 1 = t$
That is, $t = 1$
Substitute this in equation (1)
$v = 3 t^{2} - 6 t v = 3 \times 1^{2} - 6 \times 1 v = 3 - 6 v = - 3 m s^{- 1}$
Therefore, the velocity of the particle when the acceleration is zero, in $metre / \sec$ is $- 3 m s^{- 1}$ .

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If the distance ‘s’ metre traversed by a particle in t seconds is given by s=t3–3t2, then the velocity of the particle when the acceleration is zero, in metre/sec is,

If the distance ‘ $s$ ’ metre traversed by a particle in $t$ seconds is given by $s = t^{3} - 3 t^{2}$ , then the velocity of the particle when the acceleration is zero, in $metre / \sec$ is,