The particle P moves along the parabolic surface shown. When x = 0.18 m, the particle speed is v= 3.4 m/s. For this instant determine the corresponding values of r, r, 0, and 0. Both x and y are in meters. I 8 y = 7x² P
The particle P moves along the parabolic surface shown. When x = 0.18 m, the particle speed is v= 3.4 m/s. For this instant determine the corresponding values of r, r, 0, and 0. Both x and y are in meters. I 8 y = 7x² P
Elements Of Electromagnetics
7th Edition
ISBN:9780190698614
Author:Sadiku, Matthew N. O.
Publisher:Sadiku, Matthew N. O.
ChapterMA: Math Assessment
Section: Chapter Questions
Problem 1.1MA
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Transcribed Image Text:### Problem Statement
The particle \( P \) moves along the parabolic surface shown. When \( x = 0.18 \) m, the particle speed is \( v = 3.4 \) m/s. For this instant, determine the corresponding values of \( r \), \( \dot{r} \), \( \theta \), and \( \dot{\theta} \). Both \( x \) and \( y \) are in meters.
### Diagram Explanation
In the provided diagram:
- The particle \( P \) is moving along a parabolic path defined by the equation \( y = 7x^2 \).
- The diagram is plotted on an x-y coordinate system where the parabolic path opens upwards.
- The radius vector \( r \) extends from the origin \( O \) to the particle \( P \).
- The angle \( \theta \) is measured counterclockwise from the x-axis to the radius vector \( r \).
### Answers
Please fill in the values:
- \( r = \) \_\_\_\_\_ m
- \( \dot{r} = \) \_\_\_\_\_ m/s
- \( \theta = \) \_\_\_\_\_ °
- \( \dot{\theta} = \) \_\_\_\_\_ rad/s
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