Question

Use the equation for motion to answer the question.

x=x0+v0t+1/2at^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

60 m
60 m

30 m
30 m

17 m
17 m

32 m

Answer 1

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

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

Given the values:
- \( x_0 = 2 \) m (starting position)
- \( v_0 = 15 \) m/s (starting velocity)
- \( a = 0 \) m/s² (since the object is moving at a constant speed, the acceleration is zero)
- \( t = 1 \) s (time period)

Since the acceleration \( a = 0 \), the equation simplifies to:

\[
x = x_0 + v_0 t
\]
Substituting in the values:

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

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

The correct response is:

**17 m**