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

32 m
32 m

30 m
30 m

60 m
60 m

17 m

Answer 1

To calculate the final position of the object using the equation of motion \( x = x_0 + v_0 t + \frac{1}{2} a t^2 \), we need to identify the values for \( x_0 \), \( v_0 \), \( a \), and \( t \) based on the information given:

- \( x_0 = 2 \) m (starting position)
- \( v_0 = 15 \) m/s (starting velocity)
- \( a = 0 \) m/s² (constant speed means acceleration is zero)
- \( t = 1 \) s

Now we can plug these values into the equation:

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

Calculating each term:

1. \( 2 \) (initial position)
2. \( 15 \times 1 = 15 \) (distance traveled due to velocity)
3. \( \frac{1}{2} \times 0 \times 1^2 = 0 \) (no distance traveled due to acceleration)

Now we add them together:

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

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