Answer:
A. 60 mi.
B. 46 mi.
C. -14 mi.
Explanation:
A)
- Assuming that we use a coordinate system with the positive x-axis aimed toward the east, and the origin at the start point, we can find the position vector at the end of the segment 1, applying the definition of average velocity, as follows:
[tex]v_{avg1} = \frac{x_{1f} -x_{1o}}{t} (1)[/tex]
- where x₁f = final position at the end of segment 1, x₁₀ = initial position at the start of the segment 1 = 0, t = time traveled during segment 1 = 2.0 h, and v₁, average velocity during segment 1 = 30 mi/h due east.
- Replacing by the givens, and solving for x₁f, we get:
[tex]x_{1f} = v_{1avg} * t =30 mi/h * 2.0 h = 60 mi due east. (2)[/tex]
B)
- In order to find the position at the end of the segment 2, we can use the same equation (1), but taking into account that the initial position will not be zero, but the final position at the end of the segment 1, i.e., 60 mi due east.
- Replacing by the givens, and solving for x₂f, we get:
[tex]x_{2f} = x_{1f} + v_{2avg} * t = (-28 mi/h) * 0.5 h = 60 mi - 14 mi = 46 mi (3)[/tex]
C)
- The displacement during the segment two, is simply the difference between the final and initial positions for this segment.
- Since x₂₀= x₁f = 60 mi, and x₂f = 46 mi, we find that the displacement is as follows:
- Δx = x₂f - x₂₀ = 46 mi - 60 mi = - 14 mi.
