Position-time graph illustrating cart motion with labeled intervals and questions about speed, direction, and displacement.

Problem Analysis

The provided graph is a position-time graph, where the vertical axis represents position (in meters) and the horizontal axis represents time (in seconds). The problem involves analyzing the motion of a cart based on this graph. Here's a step-by-step solution to each question:

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1. During which time interval (AB, BC, CD, DE, EF, FG) was the cart traveling at its greatest speed?

- Speed is the magnitude of velocity, which is the slope of the position-time graph.
- The steeper the slope, the greater the speed.
- Observing the graph:
- AB: Slope is small and negative.
- BC: Slope is positive and moderate.
- CD: Slope is zero (horizontal line).
- DE: Slope is very steep and positive.
- EF: Slope is zero (horizontal line).
- FG: Slope is steep and negative.

The steepest slope occurs during DE. Therefore, the cart was traveling at its greatest speed during the DE interval.

Answer: DE

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2. During which time interval (AB, BC, CD, DE, EF, FG) was the cart traveling at its least (nonzero) speed?

- The least (nonzero) speed corresponds to the shallowest slope.
- Observing the graph:
- AB: Slope is small and negative.
- BC: Slope is positive and moderate.
- CD: Slope is zero (cart is at rest).
- DE: Slope is very steep and positive.
- EF: Slope is zero (cart is at rest).
- FG: Slope is steep and negative.

The shallowest nonzero slope occurs during AB.

Answer: AB

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3. During which time interval(s) (AB, BC, CD, DE, EF, FG) was the cart at rest?

- The cart is at rest when the slope of the graph is zero (i.e., the graph is horizontal).
- Observing the graph:
- CD: The graph is horizontal.
- EF: The graph is horizontal.

The cart is at rest during CD and EF.

Answer: CD and EF

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4. During which time interval(s) (AB, BC, CD, DE, EF, FG) did the cart travel in a negative direction?

- The cart travels in a negative direction when the slope of the graph is negative.
- Observing the graph:
- AB: Slope is negative.
- BC: Slope is positive.
- CD: Slope is zero.
- DE: Slope is positive.
- EF: Slope is zero.
- FG: Slope is negative.

The cart travels in a negative direction during AB and FG.

Answer: AB and FG

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5. What was the cart's instantaneous speed at 6 seconds?

- At 6 seconds, the cart is in the BC interval.
- The slope of the BC segment represents the speed.
- From the graph:
- Position at 5 seconds: Approximately \(-2\) m.
- Position at 11 seconds: Approximately \(0\) m.
- Time interval: \(11 - 5 = 6\) seconds.
- Change in position: \(0 - (-2) = 2\) m.
- Slope (speed): \(\frac{\Delta \text{position}}{\Delta \text{time}} = \frac{2}{6} \approx 0.33 \, \text{m/s}\).

Answer: 0.33 m/s

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6. What was the cart's instantaneous speed at 22 seconds?

- At 22 seconds, the cart is in the CD interval.
- The slope of the CD segment is zero (horizontal line).
- Therefore, the speed is \(0 \, \text{m/s}\).

Answer: 0 m/s

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7. What was the cart's instantaneous speed at 40 seconds?

- At 40 seconds, the cart is in the FG interval.
- The slope of the FG segment represents the speed.
- From the graph:
- Position at 40 seconds: Approximately \(8\) m.
- Position at 44 seconds: Approximately \(0\) m.
- Time interval: \(44 - 40 = 4\) seconds.
- Change in position: \(0 - 8 = -8\) m.
- Slope (speed): \(\left| \frac{\Delta \text{position}}{\Delta \text{time}} \right| = \left| \frac{-8}{4} \right| = 2 \, \text{m/s}\).

Answer: 2 m/s

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8. Describe the cart's behavior between 18 and 32 seconds. (location, speed, direction of travel, and distance traveled)

- Time Interval: 18 to 32 seconds.
- Location:
- At 18 seconds: The cart is at approximately \(-2\) m.
- At 32 seconds: The cart is at approximately \(8\) m.
- Speed:
- The slope of the DE segment is very steep and positive.
- Speed: \(\frac{\Delta \text{position}}{\Delta \text{time}} = \frac{8 - (-2)}{30 - 20} = \frac{10}{10} = 1 \, \text{m/s}\).
- Direction of Travel: Positive (North).
- Distance Traveled:
- Distance is the total change in position: \(8 - (-2) = 10 \, \text{m}\).

Answer:
- Location: From \(-2\) m to \(8\) m.
- Speed: \(1 \, \text{m/s}\).
- Direction: North.
- Distance Traveled: \(10 \, \text{m}\).

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9. What total distance did the cart travel during the entire 44 seconds?

- Distance is the total path length traveled, regardless of direction.
- Breakdown:
1. AB: From \(-5\) m to \(-2\) m → Distance = \(|-2 - (-5)| = 3 \, \text{m}\).
2. BC: From \(-2\) m to \(0\) m → Distance = \(|0 - (-2)| = 2 \, \text{m}\).
3. CD: No movement → Distance = \(0 \, \text{m}\).
4. DE: From \(0\) m to \(8\) m → Distance = \(|8 - 0| = 8 \, \text{m}\).
5. EF: No movement → Distance = \(0 \, \text{m}\).
6. FG: From \(8\) m to \(0\) m → Distance = \(|0 - 8| = 8 \, \text{m}\).
7. G: From \(0\) m to \(-2\) m → Distance = \(|-2 - 0| = 2 \, \text{m}\).

- Total distance: \(3 + 2 + 0 + 8 + 0 + 8 + 2 = 23 \, \text{m}\).

Answer: 23 m

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10. What was the cart's average speed during these 44 seconds?

- Average speed is the total distance traveled divided by the total time.
- Total distance: \(23 \, \text{m}\) (from Question 9).
- Total time: \(44 \, \text{s}\).
- Average speed: \(\frac{\text{Total Distance}}{\text{Total Time}} = \frac{23}{44} \approx 0.52 \, \text{m/s}\).

Answer: 0.52 m/s

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11. What was the cart's net displacement for the entire graph?

- Net displacement is the final position minus the initial position.
- Initial position (at \(t = 0\)): \(-5\) m.
- Final position (at \(t = 44\)): \(-2\) m.
- Net displacement: \(-2 - (-5) = 3 \, \text{m}\).

Answer: 3 m

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12. What was the cart's average velocity during these 44 seconds?

- Average velocity is the net displacement divided by the total time.
- Net displacement: \(3 \, \text{m}\) (from Question 11).
- Total time: \(44 \, \text{s}\).
- Average velocity: \(\frac{\text{Net Displacement}}{\text{Total Time}} = \frac{3}{44} \approx 0.07 \, \text{m/s}\).

Answer: 0.07 m/s

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Final Answers:

1. DE
2. AB
3. CD and EF
4. AB and FG
5. 0.33 m/s
6. 0 m/s
7. 2 m/s
8. Location: From \(-2\) m to \(8\) m; Speed: \(1 \, \text{m/s}\); Direction: North; Distance Traveled: \(10 \, \text{m}\)
9. 23 m
10. 0.52 m/s
11. 3 m
12. 0.07 m/s

Boxed Final Answer:
\[
\boxed{DE, AB, CD \text{ and } EF, AB \text{ and } FG, 0.33 \, \text{m/s}, 0 \, \text{m/s}, 2 \, \text{m/s}, \text{Location: From } -2 \, \text{m to } 8 \, \text{m; Speed: } 1 \, \text{m/s; Direction: North; Distance Traveled: } 10 \, \text{m}, 23 \, \text{m}, 0.52 \, \text{m/s}, 3 \, \text{m}, 0.07 \, \text{m/s}}
\]