To find the final position of the object after 1 second, we'll use 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
Given:
- \( x^0 = 2 , \text{m} \)
- \( v^0 = 15 , \text{m/s} \)
- \( a = 0 , \text{m/s}^2 \) (since the object is moving at a constant speed)
- \( t = 1 , \text{s} \)
Now, substitute the values into the equation:
\[ x = 2 + 15(1) + \frac{1}{2}(0)(1^2) \]
Calculating it:
\[ x = 2 + 15 + 0 \] \[ x = 17 , \text{m} \]
Therefore, the final position of the object after 1 second is \( 17 , \text{m} \).
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