Expand your horizons with the diverse and informative answers found on IDNLearn.com. Find the answers you need quickly and accurately with help from our knowledgeable and experienced experts.
Sagot :
To properly address the given problem, we shall first construct the position-time graph by plotting the provided data points [tex]\((T, X)\)[/tex]:
[tex]\[ \begin{array}{|c|c|c|c|c|c|c|c|c|c|c|c|} \hline T (s) & 0 & 5 & 10 & 15 & 20 & 25 & 30 & 35 & 90 & 95 & 50 \\ \hline X (m) & 5 & 10 & 15 & 25 & 35 & 50 & 60 & 65 & 70 & 73 & 75 \\ \hline \end{array} \][/tex]
### 1. Plotting the Position-Time Graph
To draw the graph, follow these steps:
1. On the x-axis, plot the time (T) in seconds.
2. On the y-axis, plot the position (X) in meters.
3. Mark the coordinates corresponding to each time and position pair.
4. Connect the points to visualize the motion.
### 2. Analysing Motion and its Features
Let's observe the data and the plot we would create:
#### Feature 1: Initial Velocity
- From [tex]\(T = 0\)[/tex] to [tex]\(T = 5\)[/tex]:
[tex]\[ v_1 = \frac{X(5) - X(0)}{T(5) - T(0)} = \frac{10 - 5}{5 - 0} = \frac{5}{5} = 1 \text{ m/s} \][/tex]
#### Feature 2: Consistent Speed Intervals
- From [tex]\(T = 5\)[/tex] to [tex]\(T = 10\)[/tex]:
[tex]\[ v_2 = \frac{X(10) - X(5)}{T(10) - T(5)} = \frac{15 - 10}{10 - 5} = \frac{5}{5} = 1 \text{ m/s} \][/tex]
- From [tex]\(T = 10\)[/tex] to [tex]\(T = 15\)[/tex]:
[tex]\[ v_3 = \frac{X(15) - X(10)}{T(15) - T(10)} = \frac{25 - 15}{15 - 10} = \frac{10}{5} = 2 \text{ m/s} \][/tex]
#### Feature 3: Accelerated Motion
From [tex]\(T = 15\)[/tex] to [tex]\(T = 20\)[/tex], [tex]\(T = 25\)[/tex], etc., the velocity can be observed to change significantly, suggesting acceleration:
- From [tex]\(T = 15\)[/tex] to [tex]\(T = 20\)[/tex]:
[tex]\[ v_4 = \frac{X(20) - X(15)}{T(20) - T(15)} = \frac{35 - 25}{20 - 15} = \frac{10}{5} = 2 \text{ m/s} \][/tex]
- Continued calculation for each segment up to [tex]\(T = 35\)[/tex] shows motion continues with changing velocity.
### Characteristics of Motion
1. Non-Uniform Acceleration: The body does not move with a constant acceleration throughout. There are intervals with different velocities.
2. Piecewise Linear Segments: The position-time graph has linear segments in various time intervals, indicating varying but constant velocities within those segments.
3. Final Inconsistent Motion (from [tex]\(T = 35\)[/tex] to [tex]\(T = 95\)[/tex]): There is an anomaly in recorded data at [tex]\(T=50\)[/tex], suggesting a data entry error. Otherwise, intervals from [tex]\(T = 35\)[/tex] to [tex]\(T = 95\)[/tex] show erratic motion with small increments in position.
### Conclusion
Upon detailed evaluation, this graph depicts a piecewise linear motion transitioning into a potentially erroneous record at [tex]\(T = 50\)[/tex]. The body initially moves with varying constant speeds before transitioning to a seemingly erratic pattern. Numerically significant intervals show constant speeds of 1 m/s and 2 m/s at different periods.
[tex]\[ \begin{array}{|c|c|c|c|c|c|c|c|c|c|c|c|} \hline T (s) & 0 & 5 & 10 & 15 & 20 & 25 & 30 & 35 & 90 & 95 & 50 \\ \hline X (m) & 5 & 10 & 15 & 25 & 35 & 50 & 60 & 65 & 70 & 73 & 75 \\ \hline \end{array} \][/tex]
### 1. Plotting the Position-Time Graph
To draw the graph, follow these steps:
1. On the x-axis, plot the time (T) in seconds.
2. On the y-axis, plot the position (X) in meters.
3. Mark the coordinates corresponding to each time and position pair.
4. Connect the points to visualize the motion.
### 2. Analysing Motion and its Features
Let's observe the data and the plot we would create:
#### Feature 1: Initial Velocity
- From [tex]\(T = 0\)[/tex] to [tex]\(T = 5\)[/tex]:
[tex]\[ v_1 = \frac{X(5) - X(0)}{T(5) - T(0)} = \frac{10 - 5}{5 - 0} = \frac{5}{5} = 1 \text{ m/s} \][/tex]
#### Feature 2: Consistent Speed Intervals
- From [tex]\(T = 5\)[/tex] to [tex]\(T = 10\)[/tex]:
[tex]\[ v_2 = \frac{X(10) - X(5)}{T(10) - T(5)} = \frac{15 - 10}{10 - 5} = \frac{5}{5} = 1 \text{ m/s} \][/tex]
- From [tex]\(T = 10\)[/tex] to [tex]\(T = 15\)[/tex]:
[tex]\[ v_3 = \frac{X(15) - X(10)}{T(15) - T(10)} = \frac{25 - 15}{15 - 10} = \frac{10}{5} = 2 \text{ m/s} \][/tex]
#### Feature 3: Accelerated Motion
From [tex]\(T = 15\)[/tex] to [tex]\(T = 20\)[/tex], [tex]\(T = 25\)[/tex], etc., the velocity can be observed to change significantly, suggesting acceleration:
- From [tex]\(T = 15\)[/tex] to [tex]\(T = 20\)[/tex]:
[tex]\[ v_4 = \frac{X(20) - X(15)}{T(20) - T(15)} = \frac{35 - 25}{20 - 15} = \frac{10}{5} = 2 \text{ m/s} \][/tex]
- Continued calculation for each segment up to [tex]\(T = 35\)[/tex] shows motion continues with changing velocity.
### Characteristics of Motion
1. Non-Uniform Acceleration: The body does not move with a constant acceleration throughout. There are intervals with different velocities.
2. Piecewise Linear Segments: The position-time graph has linear segments in various time intervals, indicating varying but constant velocities within those segments.
3. Final Inconsistent Motion (from [tex]\(T = 35\)[/tex] to [tex]\(T = 95\)[/tex]): There is an anomaly in recorded data at [tex]\(T=50\)[/tex], suggesting a data entry error. Otherwise, intervals from [tex]\(T = 35\)[/tex] to [tex]\(T = 95\)[/tex] show erratic motion with small increments in position.
### Conclusion
Upon detailed evaluation, this graph depicts a piecewise linear motion transitioning into a potentially erroneous record at [tex]\(T = 50\)[/tex]. The body initially moves with varying constant speeds before transitioning to a seemingly erratic pattern. Numerically significant intervals show constant speeds of 1 m/s and 2 m/s at different periods.
Thank you for using this platform to share and learn. Keep asking and answering. We appreciate every contribution you make. IDNLearn.com provides the best answers to your questions. Thank you for visiting, and come back soon for more helpful information.
