Question

Use the equation for motion to answer the question.

x=x^0+v^0^t+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?

Answer 1

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

\[
x = x^0 + v^0 t + \frac{1}{2} a t^2
\]

Where:
- \( x \) is the final position
- \( x^0 \) is the initial position
- \( v^0 \) is the initial velocity
- \( a \) is the acceleration
- \( t \) is the time

Given:
- \( x^0 = 2 \, \text{m} \)
- \( v^0 = 15 \, \text{m/s} \)
- \( a = 0 \, \text{m/s}^2 \) (since the object is moving at a constant speed)
- \( t = 1 \, \text{s} \)

Now, substitute the values into the equation:

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

Calculating it:

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

Therefore, the final position of the object after 1 second is \( 17 \, \text{m} \).