x=x0+v0t+12at2 An object has a starting position of 2 m, a starting velocity of 15 m/s, and is moving at a constant speed. Which option shows the final position of the object after 1 s? (1 point)
If an object is moving at a constant speed, the acceleration (a) is zero because it is not changing its velocity. Given this information, the term involving acceleration in the equation you provided will be zero. The equation given,
\[x = x_0 + v_0t + \frac{1}{2}at^2\]
can thus be simplified for an object moving at constant speed by removing the acceleration term:
\[x = x_0 + v_0t\]
Given the starting position (x₀) of 2 m, the starting velocity (v₀) of 15 m/s, and the fact that the object is moving at a constant speed (a = 0) for a duration (t) of 1 second, we can substitute these values into the equation:
\[x = 2\ m + (15\ m/s) \cdot (1\ s)\]
\[x = 2\ m + 15\ m\]
\[x = 17\ m\]
So, the final position of the object after 1 second is 17 meters from the starting point.