A rod of length 1 m rotates in the xy plane about the fixed point O in the anticlockwise sense, as shown in figure with velocity `omega=a+bt` where `a=10rads^(-1)` and `b=5rads^(-2)`. The velocity and acceleration of the point A at `t=0` is

Options

A. `+10hatims^(-1)` and `+5hatims^(-2)`
B. `+10hatjms^(-1)` and `(-100hati+5hatj)ms^(-2)`
C. `-10hatjms^(-1)` and `(100hati+5hatj)ms^(-2)`
D. `-10hatjms^(-1)` and `-5hatjms^(-1)`

Answer

B

Explanation

`omega=10+5t`
`alpha=(domega)/(dt)=5`
At `t=0,omega=10rad//s`
and `alpha=5rad//s^(2)impliesr=OA=1m`
`v=(romega)hatj=(10hatj)m//s`
`a=(ralpha)hatj-(romega^(2))hati`
`=(-100hati+5hatj)m//s^(2)`