Question

For a particle moving along $x$ - axis, the acceleration $a$ of the particle in terms of its $x$ - coordinate 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 maximum distance of the particle along the positive $x$-direction from origin will be

• A. $1\text{m}$
• B. $2\text{m}$
• C. $3\text{m}$
• D. $4\text{m}$

Answer 1

The correct option is B $2\text{m}$

The velocity initially is in positive direction and acceleration is in negative direction, so the velocity will keep on decreasing till the velocity becomes $0$.
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 $x=0,v=u=+6\text{m/s}$
$\Rightarrow C=18$
$\therefore \frac{v^2}{2}=-\frac{9x^2}{2}+18.....\text{(i)}$
In order to find the maximum distance travelled by the particle along positive $x$-direction, we need to put $v=0$ in eqn. $\text{(i)}$
$⟹0=-\frac{9x^2}{2}+18$
On solving, we get
$x=2\text{m}$