If velocity of the particle is given by $V = \sqrt{x}$ where ‘x’ denotes position of particle. And initially particle was at x=4, which of the following are correct

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Particle will never go in negative direction from starting position

Particles acceleration at t=2sec is $1 m / s^{2}$

at t=2sec, x=9

Particles acceleration is $\frac{1}{2} m / s^{2}$ throughout motion

answer is A, C, D.

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Detailed Solution

$V = \sqrt{x}$

$\frac{d x}{d t} = \sqrt{x} \Rightarrow \int \frac{d x}{\sqrt{x}} = \int d t$

at t=2s, ${(2 \sqrt{x})}_{4}^{x} = {(t)}_{0}^{2} \Rightarrow 2 (\sqrt{x} - \sqrt{4}) = 2$

$\sqrt{x} - 2 = 1$

X=9

$a = \frac{d v}{d t} = \frac{d v}{d x} . \frac{d x}{d t}$

$= v . \frac{1}{2 \sqrt{x}}$

$= \sqrt{x} \times \frac{1}{2 \sqrt{x}}$

$a = \frac{1}{2} m / s^{2}$

as the acceleration is always positive, the particle will not change its direction.

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