Question

If velocity of the particle is given by $v=\sqrt{x}$, where $x$ denotes the position of the particle and initially particle was at $x=4$, then which of the following are correct ?

• A. At $t=2\text{s}$, the position of the particle is $x=9$
• B. Particle's acceleration at $t=2\text{sec}$, is $1{\text{m/s}}^2$
• C. Particle's acceleration is $\frac{1}{2}{\text{m/s}}^2$ throughout the motion
• D. Particle will never go in negative direction from its starting position

Answer 1

Answer: (A), (C), (D)

The correct options are
A At $t=2\text{s}$, the position of the particle is $x=9$
C Particle's acceleration is $\frac{1}{2}{\text{m/s}}^2$ throughout the motion
D Particle will never go in negative direction from its starting position

$v=\sqrt{x}\Rightarrow \frac{dx}{dt}=\sqrt{x}\Rightarrow \frac{dx}{x^{\frac{1}{2}}}=dt\Rightarrow 2\sqrt{x}=t+C$
but given at $t=0;x=4\Rightarrow c=4$

$x=\frac{(t+4{)}^2}{4}\Rightarrow x=\frac{(6{)}^2}{4}=\frac{36}{4}=9\text{m}$ [Putting $t=2\text{sec}$]

$a=v\frac{dv}{dx}=\sqrt{x}\times \frac{1}{2\sqrt{x}}=\frac{1}{2}{\text{m/s}}^2$

And since the acceleration is always positive, the particle will never change its direction of motion.