Solved Caneousfeed POSITION-TIME GRAPH WORKSHEET 2 I at one ... - Free Printable

Problem Analysis and Solution

The provided image is a position-time graph where the vertical axis represents position (in meters) and the horizontal axis represents time (in seconds). Positive positions and directions of travel are considered to be North. We need to analyze the graph to answer the given questions.

#### Key Observations from the Graph:
1. Slope of the Line:
- The slope of the line on a position-time graph represents the velocity of the cart.
- A steeper slope indicates a higher speed.
- A positive slope means the cart is moving in the positive direction (North).
- A negative slope means the cart is moving in the negative direction (South).
- A horizontal line indicates that the cart is at rest (velocity = 0).

2. Intervals:
- AB: From 0 to 3 seconds.
- BC: From 3 to 11 seconds.
- CD: From 11 to 20 seconds.
- DE: From 20 to 30 seconds.
- EF: From 30 to 40 seconds.
- FG: From 40 to 44 seconds.

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Question-by-Question Solution

#### 1. During which time interval was the cart traveling at its greatest speed?
- The speed is determined by the slope of the line.
- The steepest slope occurs between D and E (interval DE).
- Therefore, the cart was traveling at its greatest speed during interval DE.

Answer: DE

#### 2. During which time interval was the cart traveling at its least (nonzero) speed?
- The least (nonzero) speed corresponds to the shallowest slope.
- The shallowest slope occurs between B and C (interval BC).
- Therefore, the cart was traveling at its least (nonzero) speed during interval BC.

Answer: BC

#### 3. During which time interval(s) was the cart at rest?
- The cart is at rest when the slope of the line is zero (horizontal line).
- This occurs between C and D (interval CD) and between E and F (interval EF).
- Therefore, the cart was at rest during intervals CD and EF.

Answer: CD, EF

#### 4. During which time interval(s) did the cart travel in a negative direction?
- The cart travels in a negative direction when the slope of the line is negative.
- This occurs between A and B (interval AB) and between F and G (interval FG).
- Therefore, the cart traveled in a negative direction during intervals AB and FG.

Answer: AB, FG

#### 5. What was the cart's instantaneous speed at 6 seconds?
- At 6 seconds, the cart is in the interval BC.
- The slope of the line in interval BC is calculated as follows:
\[
\text{slope} = \frac{\Delta \text{position}}{\Delta \text{time}} = \frac{-2 - (-5)}{11 - 3} = \frac{3}{8} \, \text{m/s}
\]
- Since the slope is positive, the speed is \( \frac{3}{8} \, \text{m/s} \).

Answer: \( \frac{3}{8} \, \text{m/s} \)

#### 6. What was the cart's instantaneous speed at 22 seconds?
- At 22 seconds, the cart is in the interval DE.
- The slope of the line in interval DE is calculated as follows:
\[
\text{slope} = \frac{\Delta \text{position}}{\Delta \text{time}} = \frac{8 - 2}{30 - 20} = \frac{6}{10} = 0.6 \, \text{m/s}
\]
- Since the slope is positive, the speed is \( 0.6 \, \text{m/s} \).

Answer: \( 0.6 \, \text{m/s} \)

#### 7. What was the cart's instantaneous speed at 40 seconds?
- At 40 seconds, the cart is in the interval FG.
- The slope of the line in interval FG is calculated as follows:
\[
\text{slope} = \frac{\Delta \text{position}}{\Delta \text{time}} = \frac{0 - 8}{44 - 40} = \frac{-8}{4} = -2 \, \text{m/s}
\]
- The speed is the absolute value of the slope, so the speed is \( 2 \, \text{m/s} \).

Answer: \( 2 \, \text{m/s} \)

#### 8. Describe the cart's behavior between 18 and 32 seconds.
- Between 18 and 32 seconds, the cart is in intervals CD and DE.
- CD (11 to 20 seconds): The cart is at rest (velocity = 0).
- DE (20 to 30 seconds): The cart is moving in the positive direction (North) with a constant speed.
- Location: Starts at \( x = 2 \, \text{m} \) and ends at \( x = 8 \, \text{m} \).
- Speed: 0 m/s during CD, and \( 0.6 \, \text{m/s} \) during DE.
- Direction: Stationary during CD, then moving North during DE.
- Distance Traveled: Only during DE: \( 8 - 2 = 6 \, \text{m} \).

Answer: The cart is at rest from 18 to 20 seconds, then moves North at a constant speed of \( 0.6 \, \text{m/s} \) from 20 to 32 seconds, covering a distance of 6 meters.

#### 9. What total distance did the cart travel during the entire 44 seconds?
- Distance is the total path length traveled, regardless of direction.
- Calculate the distance for each segment where the cart is moving:
- AB: From \( x = 0 \) to \( x = -5 \): Distance = \( 5 \, \text{m} \).
- BC: From \( x = -5 \) to \( x = -2 \): Distance = \( 3 \, \text{m} \).
- DE: From \( x = 2 \) to \( x = 8 \): Distance = \( 6 \, \text{m} \).
- FG: From \( x = 8 \) to \( x = 0 \): Distance = \( 8 \, \text{m} \).
- Total distance:
\[
5 + 3 + 6 + 8 = 22 \, \text{m}
\]

Answer: \( 22 \, \text{m} \)

#### 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 = \( 22 \, \text{m} \).
- Total time = \( 44 \, \text{s} \).
- Average speed:
\[
\text{Average speed} = \frac{\text{Total distance}}{\text{Total time}} = \frac{22}{44} = 0.5 \, \text{m/s}
\]

Answer: \( 0.5 \, \text{m/s} \)

#### 11. What was the cart's net displacement for the entire graph?
- Displacement is the change in position from the starting point to the ending point.
- Starting position (\( t = 0 \)): \( x = 0 \).
- Ending position (\( t = 44 \)): \( x = 0 \).
- Net displacement:
\[
\text{Displacement} = x_{\text{final}} - x_{\text{initial}} = 0 - 0 = 0 \, \text{m}
\]

Answer: \( 0 \, \text{m} \)

#### 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 = \( 0 \, \text{m} \).
- Total time = \( 44 \, \text{s} \).
- Average velocity:
\[
\text{Average velocity} = \frac{\text{Net displacement}}{\text{Total time}} = \frac{0}{44} = 0 \, \text{m/s}
\]

Answer: \( 0 \, \text{m/s} \)

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Final Answers

\[
\boxed{
\begin{array}{ll}
1. & \text{DE} \\
2. & \text{BC} \\
3. & \text{CD, EF} \\
4. & \text{AB, FG} \\
5. & \frac{3}{8} \, \text{m/s} \\
6. & 0.6 \, \text{m/s} \\
7. & 2 \, \text{m/s} \\
8. & \text{At rest from 18 to 20 seconds, then moves North at 0.6 m/s from 20 to 32 seconds, covering 6 meters.} \\
9. & 22 \, \text{m} \\
10. & 0.5 \, \text{m/s} \\
11. & 0 \, \text{m} \\
12. & 0 \, \text{m/s} \\
\end{array}
}
\]