Question

A particle is moving in such a way that at one instant velocity vector of the particle is 3î + 4j m/s and acceleration vector is -25î -25j m/s^2. The radius of curvature of the trajectory of the particle at that instant is R. Find the value of R (in meter).

Answer 1

Answer:

Explanation:

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Answer 2

Answer:

Unit vector along velocity=1/5(3i+4j)dot

Unit vector perpendicular to velocity=1/5(-3i+4j)

Component of Acceleration along normal direction=dot product of Acceleration and unit vector perpendicular to velocitiy

Or, v^2/R=5

or,R=5 m