1. Given data

• Radius of curvature of concave mirror: $R = 200\,\text{cm}$
• Hence focal length: $f = \frac{R}{2} = 100\,\text{cm}$

For a concave mirror, using Cartesian sign convention: $f = -100\,\text{cm}$

• Initial object distance: $100\,\text{cm}$ in front of mirror $u_0 = -100\,\text{cm}$

• Object moves towards the mirror with speed $2\,\text{cm/s}$ for $10\,\text{s}$

So in $10\,\text{s}$, object moves: $s = vt = 2 \times 10 = 20\,\text{cm}$

Thus new object distance is: $u = -(100 - 20) = -80\,\text{cm}$

2. Use mirror formula

Mirror formula: $\frac{1}{f} = \frac{1}{v} + \frac{1}{u}$

Substitute $f=-100\,\text{cm}$ and $u=-80\,\text{cm}$: $\frac{1}{-100} = \frac{1}{v} + \frac{1}{-80}$

$-\frac{1}{100} = \frac{1}{v} - \frac{1}{80}$

So, $\frac{1}{v} = -\frac{1}{100} + \frac{1}{80}$

Take LCM $400$: $\frac{1}{v} = \frac{-4+5}{400} = \frac{1}{400}$

Therefore, $v = 400\,\text{cm}$

3. Interpretation

The positive sign of $v$ means the image is formed behind the mirror (virtual image), at a distance of $400\,\text{cm}$ from the mirror.

4. Final answer

The position of the image after $10\,\text{s}$ is: $\boxed{400\,\text{cm}}$