Question

Use the equation for motion to answer the question.

x=x0+v0t+12at2
𝑥
=
𝑥
0
+
𝑣
0
𝑡
+
1
2
𝑎
𝑡
2

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

17 m
17 m

32 m
32 m

60 m

Answer 1

Since the object is moving at a constant speed, the acceleration (\(a\)) is 0. Thus, we can simplify the motion equation:

\[
x = x_0 + v_0 t + \frac{1}{2} a t^2
\]

Substituting \(a = 0\):

\[
x = x_0 + v_0 t
\]

Given:
- \(x_0 = 2 \, \text{m}\) (starting position)
- \(v_0 = 15 \, \text{m/s}\) (starting velocity)
- \(t = 1 \, \text{s}\) (time elapsed)

Now we can plug in the values:

\[
x = 2 \, \text{m} + (15 \, \text{m/s} \cdot 1 \, \text{s})
\]
\[
x = 2 \, \text{m} + 15 \, \text{m} = 17 \, \text{m}
\]

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