Answer

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**Hint:**Here first we have to find the arithmetic progression for $n$ and $n - 1$ seconds. And then equate the result with the given progression to get the answer.

**Complete step by step answer:**Given,

${s_n} = \left( {0.4n + 9.8} \right)\,m$

${s_n}$ distance travelled in ${n^{th}}$ and ${\left( {n - 1} \right)^{th}}$ seconds is:

Let ${u_1}$ be the initial velocity and $a$ be the acceleration

$

{s_n} = \left( {{u_1}n + \dfrac{1}

{2}a{n^2}} \right) - \left( {{u_1}\left( {n - 1} \right) + \dfrac{1}

{2}a{{\left( {n - 1} \right)}^2}} \right) \\

= {u_1} + \dfrac{1}

{2}a\left( {{n^2} - {{\left( {n - 1} \right)}^2}} \right) \\

= {u_1} + \dfrac{a}

{2}\left( { + 2n - 1} \right) \\

= {u_1} + an - \dfrac{a}

{2} \\

$

Comparing ${s_n}$ with the given equation we get,

$a = 0.4$

$

{u_1} - \dfrac{{0.4}}

{2} = 9.8 \\

{u_1} = 10\,m{s^{ - 1}} \\

$

Hence, initial velocity of the body is $10m{s^{ - 1}}$.

**Therefore option B is correct.**

**Additional information:**

Uniform motion: Uniform motion describes the motion of a body at constant speed, traversing a circular path. Since the body describes uniform motion, at all times the distance from the axis stays unchanged.

Since, the direction of the velocity varies continuously in uniform motion, so acceleration is still present but the speed may not change.

Uniform acceleration- uniform acceleration is a method of motion in which the velocity of an object varies by an equal amount in every equal time period. An instance of uniform acceleration often cited is that of an object in a uniform gravitational field in free fall.

Without the presence of any acceleration when a body is moving, the velocity applied is uniform velocity. There is no change in either direction or speed. So, acceleration is taken as zero for both uniform and constant velocity.

A uniform acceleration means that the object’s speed varies every second by the same amount. If an object’s velocity is decreasing, the measured acceleration is negative.

**Note:**Here we have to be careful while calculating the initial velocity. We cannot write the initial velocity simply as $9.8$. We have to calculate from the equation of ${n^{th}}$ velocity.

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