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**Hint:**As seen, force in itself is dependent on the position of the particle hence the acceleration is not a constant. For such a changing force, the equation of motion can be used to solve the problem, with an integral sign before the acceleration.

Formula used: In this solution we will be using the following formulae;

\[{v^2} = {u^2} + 2\int {adx} \] where \[v\] is the final velocity (or instantaneous velocity at a particular position) of a body, \[u\] is the initial velocity at the start point, \[a\] is the acceleration of the object, \[x\] signifies the distance, and \[\int {adx} \] signifies an integral of a changing acceleration with respect to the position of the object.

Acceleration \[a = \dfrac{F}{m}\] where \[F\] is the force acting on a body of mass \[m\].

**Complete Step-by-Step solution:**

The force acting on a body is in itself proportional to the position of the body. Hence, the acceleration of the body is changing. For such a situation, we have the equation of motion to be given as

\[{v^2} = {u^2} + 2\int {adx} \] where \[v\] is the final velocity of a body, \[u\] is the initial velocity at the start point, \[a\] is the acceleration of the object, \[x\] signifies the distance, and \[\int {adx} \] signifies an integral of a changing acceleration with respect to the position of the object.

But, generally, acceleration is given by

\[a = \dfrac{F}{m}\] where \[F\] is the force acting on a body of mass \[m\].

Hence,

\[{v^2} = {u^2} + 2\int {\dfrac{F}{m}dx} \]

Assuming the body begins from rest, and inserting known values and limit of integration, we have

\[{v^2} = 2\int_2^{10} {\dfrac{{3x}}{8}dx} = \dfrac{3}{4}\int_2^{10} {xdx} \]

\[ \Rightarrow {v^2} = \dfrac{3}{4}\left( {\dfrac{{{{10}^2}}}{2} - \dfrac{{{2^2}}}{2}} \right) = \dfrac{3}{8}\left( {100 - 4} \right)\]

By computation,

\[{v^2} = 36\]

\[v = 6m/s\]

**Hence, the correct option is A.**

**Note:**For understanding, the equation, \[{v^2} = {u^2} + 2\int {adx} \] can be derived from the definition of instantaneous acceleration, which is

\[a = \dfrac{{dv}}{{dt}}\]

\[ \Rightarrow dv = adt\]

But also, we know that

\[v = \dfrac{{dx}}{{dt}}\], hence,

\[dt = \dfrac{{dx}}{v}\]

Hence, inserting into \[dv = adt\], we get

\[dv = a\dfrac{{dx}}{v}\]

\[ \Rightarrow vdv = adx\]

Integrating both sides from initial to final value, we have

\[\dfrac{{{v^2} - {u^2}}}{2} = \int_{{x_o}}^{{x_1}} {adx} \]

Rearranging we have,

\[{v^2} = {u^2} + 2\int_{{x_o}}^{{x_1}} {adx} \]

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