To find the final position of the object after 2 seconds, we can use the equation of motion:
\[ x = x_0 + v_0 t + \frac{1}{2} a t^2 \]
Given:
- Initial position \( x_0 = 2 \) m
- Initial velocity \( v_0 = 4.5 \) m/s
- Acceleration \( a = 0 \) m/s² (since there is no acceleration)
- Time \( t = 2 \) s
Substituting the values into the equation:
\[ x = 2 , \text{m} + (4.5 , \text{m/s} \times 2 , \text{s}) + \frac{1}{2} (0 , \text{m/s}^2) (2 , \text{s})^2 \]
Calculating each term:
- Initial position: \( 2 , \text{m} \)
- Velocity term: \( 4.5 , \text{m/s} \times 2 , \text{s} = 9 , \text{m} \)
- Acceleration term: since \( a = 0 \), the term is \( 0 \).
Now adding these together:
\[ x = 2 , \text{m} + 9 , \text{m} + 0 = 11 , \text{m} \]
Therefore, the final position of the object after 2 seconds is 11 m.