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 $x=4m$, then which of the following are correct? (Given velocity is in $m/s$ and position is in $m$)

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

Answer 1

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

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

Given,
$v=\sqrt{x}$
$\frac{dx}{dt}=\sqrt{x}$
$\int \frac{dx}{\sqrt{x}}=\int dt$
$2\sqrt{x}=t+c$
at $t=0,x=4$
$\Rightarrow c=4$
$\Rightarrow \sqrt{x}=\frac{t}{2}+2$

Position of the particle
$x={(\frac{t}{2}+2)}^2$
$x=\frac{t^2}{4}+2t+\mathrm{4............}\left(1\right)$
$x{|}_{t=2}=1+4+4=9$

Differentiating $\left(1\right)$
$v=\frac{t}{2}+\mathrm{2................}\left(2\right)$

Differentiating $\left(2\right)$
$a=\frac{dv}{dt}$
$a=\frac{1}{2}m{s}^{-2}$ ( a is constant throughout the motion)
Since $t\ge 0\Rightarrow x\ge 0$----from$\left(1\right)$