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.
θ=2t^3-6t
θ'=dθ/dt=6t²-6
θ"=d²θ/dt²
=12t
θ(2)=12*2=24 radians-sec-2
To determine the angular acceleration when t = 2 seconds, we need to find the second derivative of the equation theta = 2t^3 - 6t with respect to time (t).
Step 1: Find the first derivative of theta with respect to t:
d(theta)/dt = d(2t^3 - 6t)/dt = 6t^2 - 6
Step 2: Find the second derivative of theta with respect to t:
d^2(theta)/dt^2 = d/dt(6t^2 - 6)
= 12t
Step 3: Substitute t = 2 into the expression for the second derivative:
d^2(theta)/dt^2 = 12(2)
= 24 radians per sec^2
Therefore, the angular acceleration when t = 2 seconds is 24 radians per sec^2.
So, the correct answer is D) 24 radians per sec^2.