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 plot of velocity versus $x$ - coordinate of particle in the duration it moves from origin towards the positive $x$ - direction, is best represented by

• A.
• B.
• C.
• D.

Answer 1

The correct option is D

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 this relation $v$ is not varying linearly with , therefore graphs (a), (b) and (c) are incorrect.
Equation $\text{(i)}$ implies quadratic relation between $v$ and $x$.
Hence, the correct answer is option (d).