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

17 m
17 m

60 m
60 m

30 m

Answer 1

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

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

Given:

- Starting position, \( x_0 = 2 \) m
- Starting velocity, \( v_0 = 15 \) m/s
- The object is moving at a constant speed, which means the acceleration \( a = 0 \) m/s².
- Time, \( t = 1 \) s

Substituting the values into the equation:

\[ x = 2 \, \text{m} + (15 \, \text{m/s})(1 \, \text{s}) + \frac{1}{2} (0 \, \text{m/s}^2)(1 \, \text{s})^2 \]

Since the acceleration term is zero, it simplifies to:

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

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

So the correct response is:

**17 m**