In the equation for motion \( x = x_0 + v_0 t + \frac{1}{2} a t^2 \):
- \( x_0 \) is the initial position
- \( v_0 \) is the initial velocity
- \( a \) is the acceleration
- \( t \) is the time
Given in the problem:
- Initial position \( x_0 = 2 \) m
- Initial velocity \( v_0 = 15 \) m/s
- Since the object is moving at a constant speed, the acceleration \( a = 0 \) m/s².
- Time \( t = 1 \) s
Now, substituting these values into the equation:
\[ x = 2 + 15(1) + \frac{1}{2}(0)(1^2) \]
This simplifies to:
\[ x = 2 + 15 + 0 \]
\[ x = 17 , \text{m} \]
Therefore, the final position of the object after 1 second is 17 m.
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