Use the equation for motion to answer the question.

Question

x=x0+v0t+12at2
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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

Answer 1

To find the final position of the object after 1 second using the equation for motion \( x = x_0 + v_0 t + \frac{1}{2} a t^2 \), we first note that the object is moving at a constant speed. This means that the acceleration \( a = 0 \).

Given:

• Starting position \( x_0 = 2 , \text{m} \)
• Starting velocity \( v_0 = 15 , \text{m/s} \)
• Time \( t = 1 , \text{s} \)

Since the acceleration is zero, the equation simplifies to:

\[ x = x_0 + v_0 t \]

Now, substituting 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.