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Acceleration of a particle in x-y plane ...

Acceleration of a particle in x-y plane varies with time as `a=(2t hati+3t^2 hatj) m//s^2` At time `t =0,` velocity of particle is `2 m//s` along positive x direction and particle starts from origin. Find velocity and coordinates of particle at `t=1s.`

A

acceleration of particle is `8 m//s^(2)`

B

tangential acceleration of particle is `(6)/(sqrt(5)) m//s^(2)`

C

radial acceleration of particle is `(2)/(sqrt(5))m//s^(2)`

D

radius of curvature to the path is `(5 sqrt(5))/(2)m`

Text Solution

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The correct Answer is:
B, C, D
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