What is the meaning of an object having an uncertainty of velocity of 2 $\rm m/s$? In several questions we are given the uncertainty in velocity of an object and are asked to calculate the uncertainty in position of an object?
Well my doubt is that,as when we say that the uncertainty in position (of an let's say electron) is let's say 2 nm
We mean that the electron could be found anywhere within that 2 nanometer it could be present in 1.2 nm Or at 1.5 nm Or anywhere within that 2nm distance
But when we say that the uncertainty in velocity (of the electron) is let's say 2m/s (just for the sake of convenience)
Then what does that literally mean as it meant in the case of position
 A: As Marko points out, this question is unclear, but I see you have the uncertainty principle as a tag.
If you mean that the electron has a velocity with uncertainty $\Delta v=2\ ms^{-1}$ and you want to find its corresponding position uncertainty, then you need to solve the following relation $$\Delta x\Delta p\ge \frac \hbar2$$ and since $p=mv$ then $\Delta p=m\Delta v$ so $$\Delta x\ge\frac{\hbar}{2m\Delta v}$$ where $m$ is the mass of the electron, then plug in the uncertainty in $v$ and Planck's constant.
A: 
In several questions we are given the uncertainty in velocity of an object and are asked to calculate the uncertainty in position of an object?

With any uncertainty, you need a function that translates probability measure space to measurable space, also called distribution of the random variable. Your example does not specify what exactly does the "uncertainty" of $2 \text{ m/s}$ mean.
Taking into account stochastic effects, the displacement is defined as
$$\Delta s = v \cdot t$$
where displacement $\Delta s$ and velocity $v$ are stochastic variables, while time $t$ is taken to be deterministic. The above equation can also be written as
$$\Delta \bar{s} + \Delta \tilde{s} = (\bar{v} + \tilde{v}) \cdot t$$
where (over)bar indicates mean (or expected) value of a stochastic variable, and tilde indicates variable uncertainty with zero mean. Since $\Delta \bar{s} = \bar{v} \cdot t$, from this it follows that the uncertainty in the displacement is
$$\Delta \tilde{s} = \tilde{v} \cdot t$$

We do not have any characterization of the velocity uncertainty (distribution). Let's discuss what happens for two common distributions: (i) uniform, and (ii) normal.
Uniform distribution
If the velocity follows uniform distribution it means that all velocity realizations lie within the $\bar{v} \pm 2 \text{ m/s}$ confidence interval, and every sample has equal chance of occurring. Since time is a scalar variable, the displacement uncertainty also follows uniform distribution and its confidence interval is $\Delta \bar{s} \pm (2 \cdot t) \text{ m}$ - again, any displacement sample from within this confidence interval has equal chance of occurring.
Normal distribution
Normal distributions are not bounded, i.e. the range of possible values goes from $-\infty$ to $+\infty$. In that sense we usually specify uncertainty via standard deviation $\sigma$ - there is 68% chance that the stochastic variable realization will be within the $\bar{x} \pm 1\sigma$ confidence interval, 95.4% within the $\bar{x} \pm 2\sigma$, 99.7% within the $\bar{x} \pm 3\sigma$ etc.
If a stochastic variable $\tilde{x}$ follows normal distribution with mean $\mu$ and standard deviation $\sigma$, written as $\tilde{x} \sim \mathcal{N}(\mu, \sigma^2)$, then $a \cdot \tilde{x}$ also follows normal distribution $a \cdot \tilde{x} \sim \mathcal{N}(a \mu, a^2 \sigma^2)$ where $a$ is (real) scalar.
In your example, if $2 \text{ m/s}$ equals one standard deviation ($1 \sigma$) of the velocity, then displacement deviation will be $(2 \cdot t) \text{ m}$, which means there is 68% chance of displacement occurring from within the $\Delta \bar{s} \pm (2 \cdot t) \text{ m}$ confidence interval.
A: When you say that electron position is uncertain by $\pm 2\,\rm nm$, it means that whatever electron position you have calculated/measured, you may be wrong by 2nm offset. That is, in reality electron could be anywhere within 2 nm radius of your target position coordinate in position vector space.
Same goes about speed uncertainty. Whatever speed of electron you have calculated/measured, $2\,\rm m/s$ uncertainty means that in $v_x,v_y,v_z$ speed vector 3D phase space, electron speed fluctuates within radius of $2\,\rm m/s$ of your target value.
In addition, each vector component, may have it's own uncertainties. That is, it can be that
$$\Delta v_x \ne \Delta v_y \ne \Delta v_z,$$
due to the fact that your particle detector may be more sensitive to some measurement axis than others, etc.
