Given that the object has a starting position (\(x_0\)) of 2 m, a starting velocity (\(v_0\)) of 4.5 m/s, and no acceleration (\(a = 0\)), we can use the equation of motion:
\[ x = x_0 + v_0 t + \frac{1}{2} a t^2 \]
In this case, since there is no acceleration, the equation simplifies to:
\[ x = x_0 + v_0 t \]
Now substituting the values:
- \(x_0 = 2 , \text{m}\)
- \(v_0 = 4.5 , \text{m/s}\)
- \(t = 2 , \text{s}\)
- \(a = 0\)
Now calculate the final position:
\[ x = 2 , \text{m} + (4.5 , \text{m/s} \times 2 , \text{s}) \] \[ x = 2 , \text{m} + 9 , \text{m} \] \[ x = 11 , \text{m} \]
So the final position of the object after 2 seconds is 11 m.