A particle moves along a straight line such that its displacement at any time $t$ is given by $s = t^{3} - 6 t^{2} + 3 t + 4$ metres. The velocity when the acceleration is zero is:-

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Answer 1

$S = t^{3} - 6^{2} + 3 t + 4$
$v = \frac{d s}{d t}$

So $= 3 t^{2} - 12 + 3 = v$
again differentiate , $a = \frac{d v}{d t}$
$6 t - 12 = 0$
$t = 2 s$
$v = 3 (2)^{2} - 12 \times 2 + 3 = - 9 m s^{- 1}$

Hence (D) is correct answer.

Answer 2

S = t^{3} - 6^{2} + 3 t + 4

v = \frac{d s}{d t}

So

= 3 t^{2} - 12 + 3 = v

again differentiate ,

a = \frac{d v}{d t}

6 t - 12 = 0

t = 2 s

v = 3 (2)^{2} - 12 \times 2 + 3 = - 9 m s^{- 1}

Hence (D) is correct answer.

A particle moves along a straight line such that its displacement at any time $t$ is given by $s = t^{3} - 6 t^{2} + 3 t + 4$ metres. The velocity when the acceleration is zero is:-

$3 m / s$

$- 12 m / s$

$42 m / s$

$- 9 m / s$

$S = t^{3} - 6^{2} + 3 t + 4$
$v = \frac{d s}{d t}$

So $= 3 t^{2} - 12 + 3 = v$
again differentiate , $a = \frac{d v}{d t}$
$6 t - 12 = 0$
$t = 2 s$
$v = 3 (2)^{2} - 12 \times 2 + 3 = - 9 m s^{- 1}$

Hence (D) is correct answer.

The position of a particle moving in the x-y plane at any time $t$ is given by : $x = (3 t^{3} - 6 t)$ metres; $y = (t^{2} - 2 t)$ metres. Select the correct statement.

Position $x$ of a particle moving along a straight line with time $t$ is given by $x = t^{2} + 2$ . Acceleration of the particle is given by.

For a particle moving along a straight line, the displacement 'x' depends on time 't' as $x = α t^{3} + β t^{2} + γ t + δ$ . The ratio of its initial acceleration to its initial velocity depends

The displacement $x$ of a particle moving along a straight line at time $t$ is given by $x = a_{0} + a_{1} t + a_{2} t^{2}$ . What is the acceleration of the particle ?

A particle moves along x-axis and its displacement at any time is given by $x (t) = 2 t^{3} - 3 t^{2} + 4 t$ in SI units. The velocity of the particle when its acceleration is zero, is:

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Average and Instantaneous Acceleration - Problem L1

A particle moves along a straight line such that its displacement at any time t is given by s=t3−6t2+3t+4 metres. The velocity when the acceleration is zero is:-

3 m/s

−12 m/s

42 m/s

−9 m/s

S=t3−62+3t+4v=dtds​ So =3t2−12+3=vagain differentiate ,a=dtdv​6t−12=0t=2sv=3(2)2−12×2+3=−9ms−1 Hence (D) is correct answer.

The position of a particle moving in the x-y plane at any time t is given by : x=(3t3−6t) metres; y=(t2−2t) metres. Select the correct statement.

Position x of a particle moving along a straight line with time t is given by x=t2+2​. Acceleration of the particle is given by.

For a particle moving along a straight line, the displacement 'x' depends on time 't' as x=αt3+βt2+γt+δ. The ratio of its initial acceleration to its initial velocity depends

The displacement x of a particle moving along a straight line at time t is given by x=a0​+a1​t+a2​t2. What is the acceleration of the particle ?

A particle moves along x-axis and its displacement at any time is given by x(t)=2t3−3t2+4t in SI units. The velocity of the particle when its acceleration is zero, is:

Revise with Concepts

Advanced Knowledge of Distance/Displacement - Time Graph

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Average and Instantaneous Acceleration - Problem L1

A particle moves along a straight line such that its displacement at any time $t$ is given by $s = t^{3} - 6 t^{2} + 3 t + 4$ metres. The velocity when the acceleration is zero is:-

$3 m / s$

$- 12 m / s$

$42 m / s$

$- 9 m / s$

$S = t^{3} - 6^{2} + 3 t + 4$
$v = \frac{d s}{d t}$

So $= 3 t^{2} - 12 + 3 = v$
again differentiate , $a = \frac{d v}{d t}$
$6 t - 12 = 0$
$t = 2 s$
$v = 3 (2)^{2} - 12 \times 2 + 3 = - 9 m s^{- 1}$

Hence (D) is correct answer.

The position of a particle moving in the x-y plane at any time $t$ is given by : $x = (3 t^{3} - 6 t)$ metres; $y = (t^{2} - 2 t)$ metres. Select the correct statement.

Position $x$ of a particle moving along a straight line with time $t$ is given by $x = t^{2} + 2$ . Acceleration of the particle is given by.

For a particle moving along a straight line, the displacement 'x' depends on time 't' as $x = α t^{3} + β t^{2} + γ t + δ$ . The ratio of its initial acceleration to its initial velocity depends

The displacement $x$ of a particle moving along a straight line at time $t$ is given by $x = a_{0} + a_{1} t + a_{2} t^{2}$ . What is the acceleration of the particle ?

A particle moves along x-axis and its displacement at any time is given by $x (t) = 2 t^{3} - 3 t^{2} + 4 t$ in SI units. The velocity of the particle when its acceleration is zero, is: