Question

6 opción múltiple 1 punto
graph: position x (m) vs time t (s), with segments a, b, c. axes: x (m) from 0 to 40, t (s) from 0 to 160. a: from (0,0) to ~(40,30); b: flat from ~40 to 80; c: decreasing from ~80 to 160.
according to the graph above, what is the objects velocity at t = 100 seconds?
options: -0.25 m/s, -0.375 m/s, -4.0 m/s, 0 m/s

Explanation:

Step1: Recall velocity from position-time graph

Velocity is the slope of the position - time graph. For a linear segment of the position - time graph, the slope \(v=\frac{\Delta x}{\Delta t}\), where \(\Delta x=x_2 - x_1\) and \(\Delta t=t_2 - t_1\). At \(t = 100\) s, the object is on the segment BC.

Step2: Identify two points on segment BC

We can take two points on segment BC. Let's take \(t_1 = 80\) s, \(x_1=30\) m and \(t_2 = 160\) s, \(x_2 = 10\) m (we can also use other points on BC, the result will be the same).

Step3: Calculate the slope (velocity)

Using the formula for slope \(v=\frac{\Delta x}{\Delta t}=\frac{x_2 - x_1}{t_2 - t_1}\). Substitute \(x_1 = 30\) m, \(x_2=10\) m, \(t_1 = 80\) s, \(t_2 = 160\) s.
\(v=\frac{10 - 30}{160 - 80}=\frac{- 20}{80}=- 0.25\) m/s? Wait, no, wait. Wait, let's check the graph again. Wait, when \(t = 80\) s, \(x = 30\) m, and when \(t=160\) s, \(x = 10\) m? Wait, no, maybe I misread the graph. Wait, the graph: at \(t = 80\) s, the position is 30 m, and at \(t = 160\) s, the position is 10 m? Wait, no, let's recalculate. Wait, the time axis: from 80 to 160 s, that's \(\Delta t=160 - 80 = 80\) s. The position axis: from 30 m to 10 m, \(\Delta x=10 - 30=- 20\) m. So \(v=\frac{-20}{80}=- 0.25\) m/s? Wait, but let's check another pair. Let's take \(t = 80\) s (\(x = 30\) m) and \(t = 120\) s. What's the position at \(t = 120\) s? From the graph, the line from \(t = 80\) s (30 m) to \(t = 160\) s (10 m). The slope is \(\frac{10 - 30}{160 - 80}=\frac{-20}{80}=- 0.25\) m/s. Wait, but let's check the options. Wait, maybe I made a mistake. Wait, let's take \(t_1 = 80\) s, \(x_1 = 30\) m and \(t_2=160\) s, \(x_2 = 10\) m. \(\Delta x=10 - 30=-20\) m, \(\Delta t=160 - 80 = 80\) s. \(v=\frac{-20}{80}=- 0.25\) m/s. Wait, but let's check the other option. Wait, maybe the two points are \(t = 80\) s (\(x = 30\) m) and \(t = 160\) s (\(x = 10\) m). Wait, no, maybe the graph is such that at \(t = 80\) s, \(x = 30\) m, and at \(t = 160\) s, \(x = 10\) m. So the velocity is \(\frac{10 - 30}{160 - 80}=\frac{-20}{80}=- 0.25\) m/s. Wait, but let's check the options. The options are - 0.25 m/s, - 0.375 m/s, - 4.0 m/s, 0 m/s. Wait, maybe I misread the graph. Wait, let's look at the time and position again. The horizontal axis: 0, 20, 40, 60, 80, 100, 120, 140, 160. The vertical axis: 0, 10, 20, 30, 40. The segment BC starts at \(t = 40\) s? No, wait, the graph: segment A is from (0,0) to (40, 30), segment B is from (40,30) to (80,30), and segment C is from (80,30) to (160,10). So for segment C, the time interval is from \(t_1 = 80\) s to \(t_2=160\) s, and the position changes from \(x_1 = 30\) m to \(x_2 = 10\) m. So \(\Delta t=160 - 80 = 80\) s, \(\Delta x=10 - 30=-20\) m. Then velocity \(v=\frac{\Delta x}{\Delta t}=\frac{-20}{80}=- 0.25\) m/s. Wait, but let's check with \(t = 80\) s and \(t = 120\) s. At \(t = 120\) s, what's the position? The slope is \(\frac{-20}{80}=- 0.25\) m/s, so from \(t = 80\) s (30 m), at \(t = 120\) s (\(t - 80=40\) s), the position is \(30+(- 0.25)\times40=30 - 10 = 20\) m. Which matches the graph (at \(t = 120\) s, the position is 20 m). So the velocity at \(t = 100\) s (which is on segment C) is - 0.25 m/s. Wait, but let's recalculate. Wait, \(\Delta t=160 - 80 = 80\) s, \(\Delta x=10 - 30=-20\) m. \(v=\frac{-20}{80}=- 0.25\) m/s. So the correct answer is - 0.25 m/s.

Wait, no, wait a second. Wait, maybe the two points are (80, 30) and (160, 10). So \(\Delta x=10 - 30=-20\), \(\Delta t=160 - 80 = 80\). So \(v=\frac{-20}{80}=- 0.25\) m/s. So the answer is - 0.25 m/s.

Answer:

-0.25 m/s