Question

For a particle moving along $x$ - axis, the acceleration $a$ of the particle in terms of its $x$- coordinate $x$ is given by $a=-9x$, where $x$ is in meters and $a$ is in ${\text{m/s}}^2$. Take acceleration, velocity and displacement in positive $x$ - direction as positive. The initial velocity of particle at $x=0$ is $u=+6\text{m/s}$. The velocity of particle at $x=2\text{m}$ will be

• A. $+6\sqrt{2}\text{m/s}$
• B. $-6\sqrt{2}\text{m/s}$
• C. $72\text{m/s}$
• D. $0$

Answer 1

The correct option is D $0$

Given $a=-9x$
We know that, $a=\frac{dv}{dt}=v\frac{dv}{dx}$
$\Rightarrow v\frac{dv}{dx}=-9x$
$\Rightarrow vdv=-9xdx$
On integrating both sides, we get
$\frac{v^2}{2}=-\frac{9x^2}{2}+C$
At $x=0,v=u=+6\text{m/s}$
$\Rightarrow C=18$
$\therefore \frac{v^2}{2}=-\frac{9x^2}{2}+18$
At $x=2\text{m}$,
$\frac{v^2}{2}=-\frac{9(2{)}^2}{2}+18$
$\Rightarrow v=0$ at $x=2\text{m}$
The answer is (d)