θ=2t^3-6t
θ'=dθ/dt=6t²-6
θ"=d²θ/dt²
=12t
θ(2)=12*2=24 radians-sec-2
A line rotates in a horizontal plane according to the equation theta=2t^3 -6t,where theta is the angular position of the rotating line, in radians ,and t is the time,in seconds. Determine the angular acceleration when t=2sec.
A.) 6 radians per sec^2
B.) 12 radians per sec^2
C.) 18 radians per sec^2
D.) 24 radians per sec^2 << my choice.
1 answer