Given the equation of motion:
\[ x = x_0 + v_0 t + \frac{1}{2} a t^2 \]
where:
- \( x_0 \) is the initial position,
- \( v_0 \) is the initial velocity,
- \( a \) is the acceleration,
- \( t \) is the time,
- \( x \) is the final position.
From the problem:
- Initial position (\( x_0 \)) = 2 m
- Initial velocity (\( v_0 \)) = 4.5 m/s
- Acceleration (\( a \)) = 0 m/s² (as there is no acceleration)
- Time (\( t \)) = 2 s
Substituting the values into the equation:
\[ x = 2 + (4.5)(2) + \frac{1}{2}(0)(2^2) \]
Calculating:
\[ x = 2 + 9 + 0 \] \[ x = 11 , \text{m} \]
Thus, the final position of the object after 2 seconds is 11 m.