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give answer in 2 step with explanation at the end of each step and final answer at the end:3) Shown below isthe graph of Y(5£ = 2.09) for a pulse travelling tothe ef witha speed of 20 ms. Draw raph of (x = ~20 1m.) 4 ] 4m 2m | 2m 4h fem ee

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give answer in 2 step with explanation at the end of each step and final answer at the end:Uploaded Image3) Shown below isthe graph of Y(5£ = 2.09) for a pulse travelling tothe ef witha speed of 20 ms. Draw raph of (x = ~20 1m.) 4 ] 4m 2m | 2m 4h fem ee

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Asked by: Firdos
Created at: 2025-11-10 03:55:34

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Let's break down the problem step-by-step: --- ## **Step 1: Understand the pulse movement and the relation between \( x \) and \( t \)** - The pulse is traveling **to the left** with a speed of \( v = 2.\, \text{m/s} \). - The given graph is \( y(x, t=2.\, \text{s}) \). - We are to find \( y(x = -2.\, \text{m}, t) \), i.e., how the displacement at \( x = -2. \) m changes with time. **Explanation:** For a pulse moving **left** at speed \( v \), the general form is \( y(x, t) = f(x + vt) \). So, at \( x = -2. \) m, \( y(-2., t) = f(-2. + 2.t) \). --- ## **Step 2: Relate the new graph to the original** - From the original graph, \( y(x, 2.) = f(x + 2. \times 2.) = f(x + 4.) \). - To relate this to \( y(-2., t) \), set \( x + 4. = -2. + 2.t \). - Therefore, when \( x = a \) in the original graph, in the new graph, \( t = \frac{a + 6}{2} \). **Explanation:** The shape of the graph will be the **same as the original**, but the x-axis values (positions) will become t-axis values (time), shifted and scaled accordingly. --- ## **Final Answer: The new graph** - The graph of \( y(x = -2.\, \text{m}, t) \) as a function of \( t \) will look **identical in shape** to the original, but the "position" values are now time values: - Where the original graph has a step from \( x = \) to \( x = 2 \), the new graph will have a step from \( t = 1 \) to \( t = 3 \) (since \( t = \frac{x + 6}{2} \)). - Similarly, map all x-values to t-values using \( t = \frac{x + 6}{2} \). **Summary Table:** | Original x (m) | New t (s) | |:--------------:|:---------:| | | 3 | | 2 | 4 | | 4 | 5 | | 6 | 6 | - So, the graph for \( y(-2., t) \) will be: - for \( t < 3 \) - 2 from \( t = 3 \) to \( t = 4 \) - from \( t = 4 \) to \( t = 5 \) - -2 linearly from \( t = 5 \) to \( t = 6 \) - for \( t > 6 \) **Sketch:** Replicate the original shape but plot it against \( t \), starting at \( t = 3 \) s. --- ### **Final Answer** The required graph of \( y(x = -2.\, \text{m}, t) \) as a function of \( t \) is a step starting at \( t = 3 \) s and ending at \( t = 6 \) s, matching the original shape but shifted and stretched according to the above mapping.

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