A particle whose velocity is given as `vecv=hati+6thatjm//s` is moving in x-y plane. At t-0, particle is at origin. Find the radius of curvature of path at point `((sqrt(2))/(3)m,(2)/(3)m)`

Options

A. 1.5 m
B. 3.0 m
C. 4.5 m
D. 6.0 m

Answer

C

Explanation

`v_(x)=1impliesx=t` and `v_(y)=6timpliesy=3t^(2)impliesy=3x^(2)`
`implies(dy)/(dx)=6x,(d^(2)y)/(dx^(2))=6implies|(dy)/(dx)|_(x)=(sqrt(2))/(3)=2sqrt(2)`
As we know that
`R=([1+((dy)/(dx))^(2)]^(3//2))/((d^(2)y)/(dx^(2)))=((1+8)^(3//2))/(6)=4.5m`