Question

velocity and acceleration quick check
use the equation for motion to answer the question.
( x = x_0 + v_0 t + \frac{1}{2} a t^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?
(1 point)
○ 17 m
○ 30 m
○ 32 m
○ 60 m

Explanation:

Step1: Identify known values

Given \( x_0 = 2 \, \text{m} \), \( v_0 = 15 \, \text{m/s} \), \( t = 1 \, \text{s} \), and constant speed (so \( a = 0 \, \text{m/s}^2 \)). The motion equation is \( x = x_0 + v_0 t + \frac{1}{2} a t^2 \).

Step2: Substitute values into the equation

Substitute \( x_0 = 2 \), \( v_0 = 15 \), \( t = 1 \), and \( a = 0 \) into the equation:
\( x = 2 + (15)(1) + \frac{1}{2}(0)(1)^2 \)
Simplify each term: \( \frac{1}{2}(0)(1)^2 = 0 \), \( (15)(1) = 15 \).
So \( x = 2 + 15 + 0 = 17 \).

Answer:

17 m