From the velocity versus time graph below, what is the acceleration of the car at the 6-second? Car's Velocity vs. Time 40 -20 -40 8. Time (seconds) 10 O m/s^2 -20 m/s^2 O +15 m/s^2 O -15 m/s^2 Velocity (m/s)
From the velocity versus time graph below, what is the acceleration of the car at the 6-second? Car's Velocity vs. Time 40 -20 -40 8. Time (seconds) 10 O m/s^2 -20 m/s^2 O +15 m/s^2 O -15 m/s^2 Velocity (m/s)
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The term "displacement" refers to when something shifts away from its original "location," and "linear" refers to a straight line. As a result, “Linear Displacement” can be described as the movement of an object in a straight line along a single axis, for example, from side to side or up and down. Non-contact sensors such as LVDTs and other linear location sensors can calculate linear displacement. Non-contact sensors such as LVDTs and other linear location sensors can calculate linear displacement. Linear displacement is usually measured in millimeters or inches and may be positive or negative.
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![**Velocity vs. Time Graph Problem**
**Question:**
From the velocity versus time graph below, what is the acceleration of the car at the 6-second mark?
**Graph Description:**
The graph provided is titled "Car’s Velocity vs. Time" and plots velocity (in meters per second) on the y-axis against time (in seconds) on the x-axis.
- **Key Points:**
- Point A (at approximately 2 seconds): Velocity is around 25 m/s.
- From A to B (approximately 4 seconds): Velocity increases, reaching around 33 m/s.
- From B to C (approximately 6 seconds): Velocity decreases to about 15 m/s.
- From C to D: Velocity again decreases steadily, reaching 0 m/s at 8 seconds.
- From D to E: Velocity decreases further to around -25 m/s.
- Finally, at point F (10 seconds): Velocity remains at this negative value.
**Calculating Acceleration:**
- The slope of the velocity-time graph represents acceleration.
- Between points B (4 seconds) and C (6 seconds), the velocity changes from approximately 33 m/s to 15 m/s.
- Acceleration is calculated as the change in velocity divided by the change in time:
\[
\text{Acceleration} = \frac{\text{Final Velocity} - \text{Initial Velocity}}{\text{Time Interval}} = \frac{15 \text{ m/s} - 33 \text{ m/s}}{6 \text{ s} - 4 \text{ s}} = \frac{-18 \text{ m/s}}{2 \text{ s}} = -9 \text{ m/s}^2
\]
**Answer Choices:**
- ○ 0 m/s²
- ○ -20 m/s²
- ○ +15 m/s²
- ○ -15 m/s²
The closest provided option to the calculated acceleration (-9 m/s²) is not available directly; however, examining the graph carefully might indicate another close option for the entire span or an error in previous analysis. The educational challenge is to verify calculations using graph slopes, reinforcing concepts of velocity changes and interpreting graphical data.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Faf89ad1c-54bc-4176-94bd-ae9c3b6c6af8%2Fab0fe436-f53d-4554-a536-921b0d3adfda%2Ft04hdo_processed.png&w=3840&q=75)
Transcribed Image Text:**Velocity vs. Time Graph Problem**
**Question:**
From the velocity versus time graph below, what is the acceleration of the car at the 6-second mark?
**Graph Description:**
The graph provided is titled "Car’s Velocity vs. Time" and plots velocity (in meters per second) on the y-axis against time (in seconds) on the x-axis.
- **Key Points:**
- Point A (at approximately 2 seconds): Velocity is around 25 m/s.
- From A to B (approximately 4 seconds): Velocity increases, reaching around 33 m/s.
- From B to C (approximately 6 seconds): Velocity decreases to about 15 m/s.
- From C to D: Velocity again decreases steadily, reaching 0 m/s at 8 seconds.
- From D to E: Velocity decreases further to around -25 m/s.
- Finally, at point F (10 seconds): Velocity remains at this negative value.
**Calculating Acceleration:**
- The slope of the velocity-time graph represents acceleration.
- Between points B (4 seconds) and C (6 seconds), the velocity changes from approximately 33 m/s to 15 m/s.
- Acceleration is calculated as the change in velocity divided by the change in time:
\[
\text{Acceleration} = \frac{\text{Final Velocity} - \text{Initial Velocity}}{\text{Time Interval}} = \frac{15 \text{ m/s} - 33 \text{ m/s}}{6 \text{ s} - 4 \text{ s}} = \frac{-18 \text{ m/s}}{2 \text{ s}} = -9 \text{ m/s}^2
\]
**Answer Choices:**
- ○ 0 m/s²
- ○ -20 m/s²
- ○ +15 m/s²
- ○ -15 m/s²
The closest provided option to the calculated acceleration (-9 m/s²) is not available directly; however, examining the graph carefully might indicate another close option for the entire span or an error in previous analysis. The educational challenge is to verify calculations using graph slopes, reinforcing concepts of velocity changes and interpreting graphical data.
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