Given the equation for the position of an object under constant acceleration:
\[ x = x_0 + v_0 t + \frac{1}{2} a t^2 \]
where:
- \( x_0 \) is the initial position,
- \( v_0 \) is the initial velocity,
- \( a \) is the acceleration,
- \( t \) is the time.
In this case:
- The initial position \( x_0 = 2 \) m,
- The initial velocity \( v_0 = 15 \) m/s,
- Since the object is moving at a constant speed, the acceleration \( a = 0 \) m/s².
Substituting the known values into the equation for time \( t = 1 \) s:
\[ x = 2 + 15(1) + \frac{1}{2} (0)(1^2) \]
Simplifying this:
\[ x = 2 + 15 + 0 \]
\[ x = 17 , \text{m} \]
Thus, the final position of the object after 1 second is 17 m.