A particle moves in a circle of radius 10 m. Its linear speed is given by v= 31 where is in second and v is in `ms^(-1)`.
(a) Find the centripetal and tangential acceleration at 1 = 2 s.
(b) Calculate the angle between the resultant acceleration and the radius vector.

Explanation

The linear speed at 1 = 2 s
`v=3t=6ms^(-1)`
The centripetal acceleration at t=2 s is
`a_(c)=(v^(2))/(r)=((6)^(2))/(10)=3.6ms^(-2)`
The tangential acceleration is `a_(t)=(dy)/(dt)=3ms^(-2)`
The angle between the radius vector with resultant acceleration is given by
`tantheta=(v^(3))/(r)=((6)^(2))/(10)=3.6ms^(-2)`
`theta=tan^(-)(0.833)=0.69` radian
In terms of degree `theta=0.69xx57.17^(@)~~40^(@)`