Question

A particle is moving along the $x$-axis. Its velocity $\left(v\right)$ with $x$ co-ordinate is varying as $v=\sqrt{x}$. Then

• A. Initial velocity of particle is zero
• B. Motion is non-uniformly accelerated
• C. Acceleration of particle at $x=2m\text{is}\frac{1}{2}m/s^2$
• D. Acceleration of particle at $x=4m\text{is}1m/s^2$

Answer 1

Answer: (A), (C)

Acceleration of particle at $x=2m\text{is}\frac{1}{2}m/s^2$

Squaring the given equation, we get
$v^2=x........\left(1\right)$
Differentiating $\left(1\right)$ w.r.t $t$, we get
$2v\frac{dv}{dt}=\frac{dx}{dt}$
$\Rightarrow a=\frac{1}{2}m/s^2$ (constant)
On comparing equation $\left(1\right)$ with $v^2=u^2+2ax,$ we get
$u=0,2a=1$
Or $a=\frac{1}{2}m/s^2$