EXERCISE 2.39 The instantaneous velocity of a point is = 10ī− 4 5 + 6 k m/s, and the acceleration is a = -30ī – 25 j+ 15 k m/s². Determine the corresponding speed, rate of change of the speed, and the radius of curvature of the path.

Elements Of Electromagnetics
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EXERCISE 2.39 The instantaneous velocity of a point is = 10 ī-4j+6 km/s, and the
acceleration is ā = -30 ī– 25 j+ 15 k m/s². Determine the corresponding speed, rate of
change of the speed, and the radius of curvature of the path.
Transcribed Image Text:EXERCISE 2.39 The instantaneous velocity of a point is = 10 ī-4j+6 km/s, and the acceleration is ā = -30 ī– 25 j+ 15 k m/s². Determine the corresponding speed, rate of change of the speed, and the radius of curvature of the path.
Expert Solution
Step 1

Solution:

Given Data:

v=10i-4j+6k m/sa=-30i-25j+15k m/s2

 

To Find:

(a) Corresponding speed.

(b) Rate of change in speed.

(c) Radius of curvature of the path.

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