The problem describes uniformly accelerated linear motion. We use the standard kinematic equations. Let the initial velocity at point A be $v_A$, the final velocity at point B be $v_B$, the time interval be $t$, the distance be $\Delta x$, and the constant acceleration be $a$.

The given values are:

$v_A = 3.0$ m/s $t = 2$ s $\Delta x = 10.0$ m

[Q1] Find the acceleration of the object. We use the kinematic equation relating displacement, initial velocity, time, and acceleration:

$$\Delta x = v_A t + \frac{1}{2} a t^2$$

To find the acceleration $a$, we rearrange the equation algebraically:

$$\Delta x - v_A t = \frac{1}{2} a t^2$$ $$a = \frac{2(\Delta x - v_A t)}{t^2}$$

Substituting the given values:

$$a = \frac{2(10.0 \text{ m} - (3.0 \text{ m/s})(2 \text{ s}))}{(2 \text{ s})^2} = \frac{2(10.0 \text{ m} - 6.0 \text{ m})}{4 \text{ s}^2} = \frac{2(4.0 \text{ m})}{4 \text{ s}^2} = 2.0 \text{ m/s}^2$$

[Q2] Find the velocity of the object as it passes point B. We use the kinematic equation for velocity as a function of time:

$$v_B = v_A + at$$

Substituting the known values, including the acceleration found in [Q1]:

$$v_B = 3.0 \text{ m/s} + (2.0 \text{ m/s}^2)(2 \text{ s}) = 3.0 \text{ m/s} + 4.0 \text{ m/s} = 7.0 \text{ m/s}$$