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**Hint:**Let us begin by understanding one-dimensional motion and instantaneous velocity. One dimensional motion applies to objects moving in straight lines. Speed is a measure of how quickly an object is moving along this line; it is a scalar quantity. Velocity is the speed with a direction, making it a vector. If an object’s velocity changes at a constant rate with time, the object is said to be accelerating. When studying motion along one dimension, there are only two possible directions for the velocity and acceleration vectors to point in.

**Formula Used:**

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

**Complete step by step solution:**

We know that velocity is the derivative of displacement.

Mathematically, we can say that \[v=\dfrac{dx}{dt}\]

Transposing sides, we can say that \[dx=v.dt\]

Integrating both sides of the above equation, we get

\[\begin{align}

& \int\limits_{0}^{x}{dx}={{v}_{0}}\int\limits_{0}^{T}{dt} \\

& \Rightarrow \left[ x \right]_{0}^{x}={{v}_{0}}\left[ t \right]_{0}^{T} \\

& \Rightarrow x={{v}_{0}}T \\

\end{align}\]

As discussed in the hint, velocity in one-dimensional motion may be positive or negative. If the velocity is negative, the displacement will be negative as well.

So, displacement cannot always take non-negative values but can vary between the negative and the positive values, like \[-{{v}_{0}}T < x < {{v}_{0}}T\].

**Hence we can say that option (B) is the correct answer.**

**Note:**We have already seen that option (A) cannot be the correct answer as the displacement can vary between positive and negative values. Similarly, acceleration is the rate of change of velocity and if the velocity is negative, the acceleration would be negative as well. Hence acceleration cannot always be a non-negative number. Whenever the velocity changes from a positive value to a negative value or vice-versa, there is a change in the direction of motion or a turning point is encountered. Hence option (D) is also incorrect.

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