An object is placed far beyond the centre of curvature of a concave...

Using the mirror formula:

$\dfrac{1}{v} + \dfrac{1}{u} = \dfrac{1}{f}$

Differentiating with respect to time $t$, we get the velocity of the image $v_i$ in terms of the velocity of the object $v_o$:

$-\dfrac{1}{v^2} \dfrac{dv}{dt} - \dfrac{1}{u^2} \dfrac{du}{dt} = 0$

$v_i = -\left(\dfrac{v}{u}\right)^2 v_o = -\dfrac{f^2}{(u-f)^2} v_o$

Since the object is moving towards the mirror, $u$ is increasing (becoming less negative), so $v_o > 0$. This makes $v_i < 0$, meaning the image moves in the negative direction (away from the mirror). For an object moving from $-\infty$ to the centre of curvature $C$, the image moves from the focus $F$ to $C$. Thus, the image moves towards the centre of curvature.

To find the acceleration of the image $a_i$, we differentiate $v_i$ with respect to $t$:

$a_i = \dfrac{dv_i}{dt} = -\dfrac{f^2}{(u-f)^2} a_o + \dfrac{2f^2}{(u-f)^3} v_o^2$

Here, $a_o > 0$ because the object is accelerating towards the mirror. Since the object is placed in front of the concave mirror, $u$ and $f$ are negative, and $|u| > |f|$, which means $(u-f) < 0$.

Therefore, both terms in the expression for $a_i$ are negative, yielding $a_i < 0$.

Since both the velocity $v_i$ and the acceleration $a_i$ of the image are negative, the speed of the image is increasing in the direction away from the mirror. Hence, the image accelerates towards the centre of curvature.

Answer: The image will accelerate towards the centre of curvature.