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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)`A. 1.5 mB. 3.0 mC. 4.5 mD. 6.0 m |
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Answer» Correct Answer - C `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` |
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