2. refer to the position-time graph for an object moving west. create a corresponding velocity-time graph…

Explanation:

Step1: Identify time intervals

First, split the position-time graph into distinct time segments:

1. $0\ \text{s}$ to $4\ \text{s}$
2. $4\ \text{s}$ to $6\ \text{s}$
3. $6\ \text{s}$ to $8\ \text{s}$
4. $8\ \text{s}$ to $12\ \text{s}$

Step2: Calculate velocity for each segment

Velocity $v = \frac{\Delta x}{\Delta t}$, where $\Delta x$ is position change, $\Delta t$ is time change. Position decreases (west direction, negative velocity):

1. 0 to 4 s: $\Delta x = 4\ \text{m} - 0\ \text{m} = 4\ \text{m}$, $\Delta t = 4\ \text{s} - 0\ \text{s} = 4\ \text{s}$ $v = \frac{4}{4} = 1\ \text{m/s west} = -1\ \text{m/s}$
2. 4 to 6 s: $\Delta x = 6\ \text{m} - 4\ \text{m} = 2\ \text{m}$, $\Delta t = 6\ \text{s} - 4\ \text{s} = 2\ \text{s}$ $v = \frac{2}{2} = 1\ \text{m/s west} = -1\ \text{m/s}$
3. 6 to 8 s: $\Delta x = 6\ \text{m} - 6\ \text{m} = 0\ \text{m}$, $\Delta t = 8\ \text{s} - 6\ \text{s} = 2\ \text{s}$ $v = \frac{0}{2} = 0\ \text{m/s}$
4. 8 to 12 s: $\Delta x = 12\ \text{m} - 6\ \text{m} = 6\ \text{m}$, $\Delta t = 12\ \text{s} - 8\ \text{s} = 4\ \text{s}$ $v = \frac{6}{4} = 1.5\ \text{m/s west} = -1.5\ \text{m/s}$

Step3: Map to velocity-time graph

Plot constant velocity values for each time interval:

• 0-4 s: $v=-1\ \text{m/s}$
• 4-6 s: $v=-1\ \text{m/s}$
• 6-8 s: $v=0\ \text{m/s}$
• 8-12 s: $v=-1.5\ \text{m/s}$

Answer:

The corresponding velocity-time graph has the following horizontal segments (with velocity plotted on the y-axis, time on the x-axis):

1. A horizontal line at $v=-1\ \text{m/s}$ from $t=0\ \text{s}$ to $t=6\ \text{s}$
2. A horizontal line at $v=0\ \text{m/s}$ from $t=6\ \text{s}$ to $t=8\ \text{s}$
3. A horizontal line at $v=-1.5\ \text{m/s}$ from $t=8\ \text{s}$ to $t=12\ \text{s}$

(Note: The negative sign indicates motion toward the west; if labeling with direction instead of sign, use "1 m/s west", "0 m/s", and "1.5 m/s west" for each interval.)