To find the final position of the object using the given equation of motion \( x = x_0 + v_0t + \frac{1}{2}at^2 \), we can substitute the known values:
- \( x_0 = 2 , \text{m} \) (starting position)
- \( v_0 = 4.5 , \text{m/s} \) (starting velocity)
- \( a = 0 , \text{m/s}^2 \) (no acceleration)
- \( t = 2 , \text{s} \) (time)
Now plug these values into the equation:
\[ x = 2 , \text{m} + (4.5 , \text{m/s} \times 2 , \text{s}) + \frac{1}{2}(0)(2 , \text{s})^2 \]
This simplifies to:
\[ x = 2 , \text{m} + (4.5 \times 2) + 0 \] \[ x = 2 , \text{m} + 9 , \text{m} \] \[ x = 11 , \text{m} \]
Therefore, the final position of the object after 2 seconds is 11 m.