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give answer in 2 step with explanation at the end of each step and final answer at the end:Se = de HERE § Erp pri SA =

Question:

give answer in 2 step with explanation at the end of each step and final answer at the end:Uploaded ImageSe = de HERE § Erp pri SA =

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Asked by: Rahul
Created at: 2025-12-05 05:29:33

Final Answer:

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Let's break down **part (a)** of the problem in **2 steps**: Draw a position (vs. time) and velocity (vs. time) graph for the described motion.* --- ## **Step 1: Analyze and Sketch Position vs. Time** ### **Given Motions:** 1. Begins 2 m above the origin. 2. Moves down 4 m in the next 2 minutes. 3. Moves down 16 m in the next 3 minutes. 4. Remains at rest for 1 minute. 5. Moves up 10 m in 2 minutes. 6. Remains at rest for 10 minutes. ### **Position vs. Time (x-t) Graph:** - **Start:** At \( x = 2 \) m. - **From to 2 min:** Move down 4 m, so \( 2 - 4 = -2 \) m (linear decrease). - **From 2 to 5 min:** Move down 16 m, \( -2 - 16 = -18 \) m (steeper linear decrease). - **From 5 to 6 min:** Rest, so position stays at \( -18 \) m (horizontal line). - **From 6 to 8 min:** Move up 10 m, \( -18 + 10 = -8 \) m (linear increase). - **From 8 to 18 min:** Rest, position stays at \( -8 \) m (horizontal line). **Explanation:** Each segment is straight (constant velocity, zero acceleration), and horizontal where the person is at rest. --- ## **Step 2: Analyze and Sketch Velocity vs. Time** ### **Velocity for Each Segment:** 1. ** to 2 min:** \[ \text{Velocity} = \frac{-4\,m}{2\,min} = -2\,\text{m/min} \] 2. **2 to 5 min:** \[ \text{Velocity} = \frac{-16\,m}{3\,min} \approx -5.33\,\text{m/min} \] 3. **5 to 6 min:** \[ \text{Velocity} = \,\text{m/min} \] 4. **6 to 8 min:** \[ \text{Velocity} = \frac{+10\,m}{2\,min} = +5\,\text{m/min} \] 5. **8 to 18 min:** \[ \text{Velocity} = \,\text{m/min} \] **Explanation:** The velocity is constant for each segment (matching the slope of the position vs. time graph), and zero when the person is at rest. --- ## **Final Answer:** ### **Position vs. Time (x-t):** - Starts at 2 m. - Decreases linearly to -2 m in 2 min. - Decreases more steeply to -18 m by 5 min. - Stays flat at -18 m until 6 min. - Increases linearly to -8 m by 8 min. - Stays flat at -8 m until 18 min. ### **Velocity vs. Time (v-t):** - \(-2\,\text{m/min}\) from –2 min. - \(-5.33\,\text{m/min}\) from 2–5 min. - \(\,\text{m/min}\) from 5–6 min. - \(+5\,\text{m/min}\) from 6–8 min. - \(\,\text{m/min}\) from 8–18 min.

Question:

All parts solved please

This is AI generated Response.
Asked by: Rahul
Created at: 2025-12-05 05:30:12

Final Answer:

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# Motion Analysis of an Object ## **Step 1: Position vs. Time (x-t) Graph** ### **Given Motion Segments:** 1. **Initial Position:** Starts at \( x = 2 \) m. 2. **First Motion:** Moves down 4 m in the next 2 minutes. 3. **Second Motion:** Moves down 16 m in the next 3 minutes. 4. **Rest Phase:** Remains at rest for 1 minute. 5. **Third Motion:** Moves up 10 m in 2 minutes. 6. **Final Rest Phase:** Remains at rest for 10 minutes. ### **Position Calculation:** - **From 0 to 2 min:** \( 2 - 4 = -2 \) m (linear decrease). - **From 2 to 5 min:** \( -2 - 16 = -18 \) m (steeper linear decrease). - **From 5 to 6 min:** Remains at \( -18 \) m (horizontal line). - **From 6 to 8 min:** \( -18 + 10 = -8 \) m (linear increase). - **From 8 to 18 min:** Remains at \( -8 \) m (horizontal line). **Explanation:** The position graph consists of linear segments representing the object's motion and horizontal segments during rest. --- ## **Step 2: Velocity vs. Time (v-t) Graph** ### **Velocity Calculation for Each Segment:** 1. **From 0 to 2 min:** \[ \text{Velocity} = \frac{-4\,m}{2\,min} = -2\,\text{m/min} \] 2. **From 2 to 5 min:** \[ \text{Velocity} = \frac{-16\,m}{3\,min} \approx -5.33\,\text{m/min} \] 3. **From 5 to 6 min:** \[ \text{Velocity} = 0\,\text{m/min} \quad (\text{at rest}) \] 4. **From 6 to 8 min:** \[ \text{Velocity} = \frac{+10\,m}{2\,min} = +5\,\text{m/min} \] 5. **From 8 to 18 min:** \[ \text{Velocity} = 0\,\text{m/min} \quad (\text{at rest}) \] **Explanation:** The velocity graph reflects constant velocities during motion periods and zero velocity during rest. --- ## **Final Answer Summary:** ### **Position vs. Time (x-t):** - **Starts at:** \( 2 \) m - **Decreases to:** \( -2 \) m by 2 min - **Decreases to:** \( -18 \) m by 5 min - **Remains at:** \( -18 \) m until 6 min - **Increases to:** \( -8 \) m by 8 min - **Remains at:** \( -8 \) m until 18 min ### **Velocity vs. Time (v-t):** - **From 0 to 2 min:** \( -2\,\text{m/min} \) - **From 2 to 5 min:** \( -5.33\,\text{m/min} \) - **From 5 to 6 min:** \( 0\,\text{m/min} \) - **From 6 to 8 min:** \( +5\,\text{m/min} \) - **From 8 to 18 min:** \( 0\,\text{m/min} \) This completes the analysis of the motion.

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