In one dimensional motion, instantaneous speed v satisfies 0 ≤ v < v0.

The correct option is: The displacement x in time T satisfies – vo T < x < vo T

Step-by-Step Explanation:

1. Understanding Instantaneous Speed:
We are given that the instantaneous speed $v$ of a particle in one-dimensional motion satisfies:
$0\le v<v_0$
where $v_0$ is a positive constant representing the strict upper limit of the speed.

2. Relating Speed to Velocity:
In one-dimensional motion along the x-axis, the velocity $v_x$ can be positive or negative depending on the direction of motion. Since speed is the magnitude of velocity ($v=\left|v_x\right|$), the condition $0\le v<v_0$ implies that:
$-v_0<v_x<v_0$

3. Calculating Displacement:
Displacement $x$ in a time interval $T$ (from $t=0$ to $t=T$) is given by the definite integral of velocity with respect to time:
$x={\int }_0^T{v}_{x}dt$

4. Applying Inequalities to the Integral:
Since $-v_0<v_x<v_0$ holds true at every instant throughout the time interval $T$, we can integrate these limits over the time interval from $0$ to $T$:
${\int }_0^T(-v_0)dt<{\int }_0^T{v}_{x}dt<{\int }_0^T{v}_{0}dt$
Evaluating the leftmost and rightmost integrals (since $v_0$ is a constant):
$-v_{0}T<x<v_{0}T$

5. Conclusion:
This confirms that the displacement $x$ must strictly lie within the range $-v_{0}T<x<v_{0}T$. Therefore, the second option is correct.