1. Given motion

The particle moves as $x(t)=x_0\sin^2\left(\frac{t}{2}\right), \qquad x_0=1\text{ m}.$

Use the identity $\sin^2\left(\frac{t}{2}\right)=\frac{1-\cos t}{2}.$ So, $x=\frac{x_0}{2}(1-\cos t).$

Hence, $\cos t=1-\frac{2x}{x_0}.$

2. Find velocity

Differentiate $x(t)$: $v=\frac{dx}{dt}=x_0\cdot 2\sin\left(\frac{t}{2}\right)\cos\left(\frac{t}{2}\right)\cdot \frac12.$ Thus, $v=x_0\sin\left(\frac{t}{2}\right)\cos\left(\frac{t}{2}\right)=\frac{x_0}{2}\sin t.$

Therefore, $v^2=\frac{x_0^2}{4}\sin^2 t.$

Now, $\sin^2 t=1-\cos^2 t=1-\left(1-\frac{2x}{x_0}\right)^2.$

So, $v^2=\frac{x_0^2}{4}\left[1-\left(1-\frac{2x}{x_0}\right)^2\right].$

Expand: $\left(1-\frac{2x}{x_0}\right)^2=1-\frac{4x}{x_0}+\frac{4x^2}{x_0^2}.$ Hence, $1-\left(1-\frac{2x}{x_0}\right)^2=\frac{4x}{x_0}-\frac{4x^2}{x_0^2}.$

Therefore, $v^2=\frac{x_0^2}{4}\left(\frac{4x}{x_0}-\frac{4x^2}{x_0^2}\right)=x_0x-x^2.$

3. Kinetic energy as a function of $x$

$K=\frac12 mv^2=\frac12 m(x_0x-x^2).$

Since $x_0=1$, $K=\frac12 m(x-x^2).$

This is a quadratic in $x$:

• opens downward,
• zero at $x=0$ and $x=1$,
• maximum at $x=\frac{x_0}{2}=\frac12.$

So the correct graph is an inverted parabola between $x=0$ and $x=1$, touching the axis at both ends.

4. Match with options

That corresponds to Option D.

5. Comparison with stored answer

Stored correct answer: D

My derived answer: D

They agree.