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:

2.5 m s^{- 1}

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3.5 m s^{- 1}

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4.54 m s^{- 1}

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8.5 m s^{- 1}

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Solution

The correct option is A $2.5 m s^{- 1}$
Given $x (t) = 2 t^{3} - 3 t^{2} + 4 t$

So, velocity $v = \frac{d x}{d t} = (6 t^{2} - 6 t + 4)$
and acceleration $a = \frac{d v}{d t} = (12 t - 6)$

When acceleration is zero ie., $(12 t - 6) = 0$
or $t = \frac{6}{12} = \frac{1}{2} s$

$∴$ Velocity of the particle at zero acceleration is

$v = 6 {(\frac{1}{2})}^{2} - 6 (\frac{1}{2}) + 4 = 2.5 m / s$

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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:

The correct option is A 2.5 ms−1Given x(t)=2t3−3t2+4tSo, velocity v=dxdt=(6t2−6t+4)and acceleration a=dvdt=(12t−6)When acceleration is zero ie., (12t−6)=0or t=612=12 s∴ Velocity of the particle at zero acceleration isv=6(12)2−6(12)+4=2.5 m/s

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: