Question

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

17 m
17 m

30 m
30 m

60 m
60 m

32 m
32 m
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Answer 1

To find the final position \( x \) of the object after 1 second, we can use the equation for motion:

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

Given that the starting position \( x_0 = 2 \) m, the starting velocity \( v_0 = 15 \) m/s, and the object is moving at a constant speed, we can assume that the acceleration \( a = 0 \) m/s² (since constant speed implies no acceleration).

Substituting the values into the equation:

1. Since \( a = 0 \), the equation simplifies to:
\[
x = x_0 + v_0 t
\]

2. Plugging in the known values:
\[
x = 2 \, \text{m} + (15 \, \text{m/s} \times 1 \, \text{s})
\]

3. This gives us:
\[
x = 2 \, \text{m} + 15 \, \text{m} = 17 \, \text{m}
\]

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