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.

A particle moves in a circle of radius 10 m. Its linear speed is given

1.	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.
Answer» SOLUTION :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^(@)`