\(\omega=0+2(20)=40(\) according to rotational equation \(\omega=\omega 0+a t)\)
and \(\theta=0+1 / 2(2)(20) 2=400\) radian
in middle uniform part angle rotated \(=40 \times 10=400\) radian
and since acceleration in first part and retardation in last part are equal so angle rotated in last part \(=400\) radian
total \(=1200\) radian