Question

Use the equation for motion to answer the question.

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)
Responses

30 m
30 m

32 m
32 m

17 m
17 m

60 m

Answer 1

To find the final position of the object after 1 second, we can use the equation for motion given:

\[ x = x_0 + v_0t + \frac{1}{2}at^2 \]

In this case:
- \( x_0 = 2 \) m (initial position)
- \( v_0 = 15 \) m/s (initial velocity)
- \( a = 0 \) m/s² (since it's moving at a constant speed, the acceleration is zero)
- \( t = 1 \) s (time)

Substituting these values into the equation, we get:

\[
x = 2 + (15)(1) + \frac{1}{2}(0)(1^2)
\]

This simplifies to:

\[
x = 2 + 15 + 0
\]
\[
x = 17 \text{ m}
\]

Hence, the final position of the object after 1 second is **17 m**.