Answer

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**Hint:**Here we have to first use the average velocity and distance to find the time. Then we can use the first equation of motion to find the acceleration.In physics, equations of motion are classified as equations that characterise a physical system’s behaviour in terms of its motion as a function of time.

**Complete step by step answer:**

For deriving components such as displacement, velocity, time and acceleration, there are three motion equations that can be used.

The first equation of motion deals with final velocity, initial velocity, acceleration and time.

Given,

Average velocity,${V_{avg}} = 0.34\,m{s^{ - 1}}$. Distance,$D = 3.06\,m$

We know that,

$

{V_{avg}} = \dfrac{D}{t} \\

\Rightarrow 0.34 = \dfrac{{3.06}}{t} \\

\Rightarrow t = 9\,\sec \\ $

We know that,

First equation of motion is given by:

$

v = u + at \\

\Rightarrow at = v - u \\

\Rightarrow 0.18 = a \times 9 \\

\therefore a = 0.02\,m{s^{ - 2}} \\

$

**Hence, option B is correct.**

**Additional information:**

Acceleration is known to be used as a term for speeding up and deceleration for speeding down. However, speeding up may be referred to as positive acceleration and slowing down as negative acceleration.The average velocity of the object is the total displacement of the object divided by the total time taken.

In other words, the pace at which an object changes its location from one place to another. Average velocity is a quantity of vectors. The SI unit is one metre per second. The sum of the original and final velocity is separated by $2$ in order to find the average. The average velocity calculator uses a formula that indicates the average velocity equal to the sum of the final velocity and the original velocity over $2$.

**Note:**Here since, the distance is given, so we have used average velocity as distance divided by time. Here we have to check the units first whether they are in SI or not. Also, we can get easily confused between the options as the difference between the options is only $0.01$. Hence, we have to be careful while calculating the answer.

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