The position [i]vs.[/i] time graph for an object moving in a straight line is shown below. What is the instantaneous velocity at $\text{t = 2 s}$?
[asy]
size(300);
// x-axis
draw((0,0)--(12,0));
// x-axis tick marks
draw((4,0)--(4,-0.2));
draw((8,0)--(8,-0.2));
draw((12,0)--(12,-0.2));
// x-axis labels
label("1",(4,0),2*S);
label("2",(8,0),2*S);
label("3",(12,0),2*S);
//y-axis
draw((0,-3)--(0,5));
//y-axis tick marks
draw((-0.2,-2)--(0,-2));
draw((-0.2,0)--(0,0));
draw((-0.2,2)--(0,2));
draw((-0.2,4)--(0,4));
// y-axis labels
label("-2",(0,-2),2*W);
label("0",(0,0),2*W);
label("2",(0,2),2*W);
label("4",(0,4),2*W);
// Axis Labels
label("Time (s)",(10,0),N);
label(rotate(90)*"Position (m)",(0,1),6*W);
//Line
draw((0,4)--(12,-2));
[/asy]

(a) $-\text{2 m/s}$
(b) $-\frac{1}{2} \ \text{m/s}$
(c) $\text{0 m/s}$
(d) $\text{2 m/s}$
(e) $\text{4 m/s}$
To find the instantaneous velocity at \( t = 2 \) seconds, we need to determine the slope of the position vs. time graph at that specific point. The slope of the position vs. time graph gives us the velocity.

1. **Identify the points on the graph:**
   - The graph is a straight line from \( (0, 4) \) to \( (12, -2) \).

2. **Calculate the slope of the line:**
   - The slope \( m \) of a line passing through points \( (x_1, y_1) \) and \( (x_2, y_2) \) is given by:
     \[
     m = \frac{y_2 - y_1}{x_2 - x_1}
     \]
   - Here, \( (x_1, y_1) = (0, 4) \) and \( (x_2, y_2) = (12, -2) \).

3. **Substitute the points into the slope formula:**
   \[
   m = \frac{-2 - 4}{12 - 0} = \frac{-6}{12} = -\frac{1}{2}
   \]

4. **Interpret the slope:**
   - The slope of the position vs. time graph is \(-\frac{1}{2}\), which represents the instantaneous velocity of the object.

5. **Verify the time \( t = 2 \) seconds:**
   - Since the graph is a straight line, the slope (and thus the velocity) is constant throughout. Therefore, the instantaneous velocity at \( t = 2 \) seconds is the same as the slope of the line.

Conclusion:
The instantaneous velocity at \( t = 2 \) seconds is \(-\frac{1}{2} \ \text{m/s}\).

The final answer is \( \boxed{ -\frac{1}{2} \ \text{m/s} } \).