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**Hint:**In order to find speed of the body at midpoint C we will use the Newton’s equation of motion which are ${v^2} - {u^2} = 2aS$ where $v$ is the final velocity , $u$ is the initial velocity and $a$ is the acceleration of the body in covering a distance of $S$.

**Complete answer:**

It’s given that C is the midpoint Of AB so, let

$AC = CB = x$ And $AB = 2x$

Let uniform acceleration of the body is $a$ and while going from point A to point B

Initial velocity is $u$

Final velocity is $v$

Distance is $2x$

Using, ${v^2} - {u^2} = 2aS$

We get, ${v^2} - {u^2} = 4ax \to (i)$

Now, suppose the velocity of the body at point C is ${V_C}$ then while going from point A to point C:

Initial velocity is $u$

Final velocity is ${V_C}$

Distance is $x$

Using ${v^2} - {u^2} = 2aS$ we get,

${V_C}^2 - {u^2} = 2ax$

From equation $(i)$ put $ax = \dfrac{{{v^2} - {u^2}}}{4}$ in ${V_C}^2 - {u^2} = 2ax$

We get,

${V_C}^2 - {u^2} = \dfrac{{{v^2} - {u^2}}}{2}$

On rearranging terms we get,

${V_C}^2 = \dfrac{{{v^2} + {u^2}}}{2}$

${V_C} = \sqrt {\dfrac{{{v^2} + {u^2}}}{2}} $

Hence, the velocity of the body at midpoint of AB at C is ${V_C} = \sqrt {\dfrac{{{v^2} + {u^2}}}{2}} $

**Note:**It should be remembered that, acceleration is uniformly constant so at any point of journey its value and direction will remain same and other two equations of motion by newton is written as $v = u + at$ and last one is $S = ut + \dfrac{1}{2}a{t^2}$ , these three equations of motion almost describe whole Kinematics.

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