Question

Question
Use the equation of motion to answer the question.

x=x0+v0t+12at2


An object has a starting position of x = 2 m, a starting velocity of 4.5 m/s, and no acceleration. Which option shows the final position of the object after 2 s?

(1 point)
Responses

5.5 m
5.5 m

18 m
18 m

9 m
9 m

11 m
11 m
Skip to navigation

Answer 1

To find the final position of the object using the given equation of motion \( x = x_0 + v_0t + \frac{1}{2}at^2 \), we can substitute the known values:

- \( x_0 = 2 \, \text{m} \) (starting position)
- \( v_0 = 4.5 \, \text{m/s} \) (starting velocity)
- \( a = 0 \, \text{m/s}^2 \) (no acceleration)
- \( t = 2 \, \text{s} \) (time)

Now plug these values into the equation:

\[
x = 2 \, \text{m} + (4.5 \, \text{m/s} \times 2 \, \text{s}) + \frac{1}{2}(0)(2 \, \text{s})^2
\]

This simplifies to:

\[
x = 2 \, \text{m} + (4.5 \times 2) + 0
\]
\[
x = 2 \, \text{m} + 9 \, \text{m}
\]
\[
x = 11 \, \text{m}
\]

Therefore, the final position of the object after 2 seconds is **11 m**.