1. Given condition

The particle moves in a straight line with retardation proportional to displacement.

So acceleration is opposite to displacement and we can write

$a=-kx$

where $k>0$ is a constant.

2. Use the kinematic relation

We use

$a=v\frac{dv}{dx}$

Hence,

$v\frac{dv}{dx}=-kx$

or

$v\,dv=-k x\,dx$

3. Integrate

Integrating,

$\int v\,dv=-k\int x\,dx$

$\frac{v^2}{2}=-\frac{kx^2}{2}+C$

Multiplying by mass $m$,

$\frac{1}{2}mv^2 = -\frac{1}{2}mkx^2 + C'$

But $\frac{1}{2}mv^2$ is the kinetic energy $K$.

So,

$K = C' - \frac{1}{2}mkx^2$

4. Find loss of kinetic energy

If initial kinetic energy is $K_0=C'$, then after displacement $x$,

$K_0-K=\frac{1}{2}mkx^2$

Thus the loss of kinetic energy is proportional to

$x^2$

5. Check options

• A: $x$ ❌
• B: $e^x$ ❌
• C: $x^2$ ✅
• D: $\log_e x$ ❌

Therefore, the correct option is C.