Given the equation of motion:
\[ x = x_0 + v_0 t + \frac{1}{2} a t^2 \]
where:
- \(x\) is the final position,
- \(x_0\) is the initial position,
- \(v_0\) is the initial velocity,
- \(a\) is the acceleration,
- \(t\) is the time.
In this case:
- \(x_0 = 2 , \text{m}\)
- \(v_0 = 4.5 , \text{m/s}\)
- \(a = 0 , \text{m/s}^2\) (no acceleration)
- \(t = 2 , \text{s}\)
Since the acceleration is zero, the equation simplifies to:
\[ x = x_0 + v_0 t \]
Substituting in the values:
\[ x = 2 , \text{m} + (4.5 , \text{m/s} \times 2 , \text{s}) \] \[ x = 2 , \text{m} + 9 , \text{m} \] \[ x = 11 , \text{m} \]
Thus, the final position of the object after 2 seconds is 11 m.