A wheel starting from rest is uniformly accelerated at $2rad/{s^2}$for 20 seconds. It is allowed to rotate uniformly for the next 10 seconds and is finally brought to rest in the next 20 seconds. The total angle stated by the wheels (in radian) is
A. 600
B. 1200
C. 1800
D. 300

Hint: Rotational motion is defined as the motion of an object around a circular path. As we are aware of the fact that translational motion and rotational motion are analogous to each other. So, to find the solution of the given question we will use the kinematic equations of rotational motion.

Formula Used: $\omega = {\omega _ \circ } + at$

Complete step by step solution:
The terms we use in translational motion such as velocity and acceleration are analogous to the term’s angular velocity and angular acceleration in rotational motion. In that respect, we see that the linear motion of a body in translational motion is analogous to the rotation of a body about a fixed axis.
According to the equation of rotational equation we get,
$\omega = {\omega _ \circ } + at$
As it is given that initially the wheel is at rest, i.e. ${\omega _ \circ } = 0$
So, $\omega = 0 + 2\left( {20} \right) = 40$
Now, distance covered by the wheel is given as,
${\theta _1} = 0 + \dfrac{1}{2} \times 2 \times {\left( {30} \right)^2} = 400radian$
Now, it rotates uniformly for 10 seconds with $\omega = 40rad/s$
So, ${\theta _2} = 40 \times 10 = 400radian$
Similarly, when brought to rest after 20 seconds = 400 radian Since, the acceleration in the first part and the retardation n the last part is same.
$\therefore $Total angle rotated by the wheel = 1200 radian
Thus, the total angle stated by the wheels (in radian) is 1200 radian.

Hence, option (B) is the correct answer.

Note:
We see that the kinematic quantities in rotational motion are angular displacement ‘$\theta $’, angular velocity ‘$\omega $’, and angular acceleration ‘$\alpha $’, which is analogous to the displacement ‘x’, velocity ‘v’, and acceleration ‘a’ in a linear motion.
So, the kinematic equation of rotational motion is given as,
$\eqalign{
  & \omega = {\omega _ \circ } + \alpha t \cr
  & \theta - {\theta _ \circ } = {\omega _ \circ } + \dfrac{1}{2}\alpha {t^2} \cr
  & {\omega ^2} = {\omega _ \circ }^2 + 2\alpha \left( {\theta - {\theta _ \circ }} \right) \cr} $
Where ‘${\omega _ \circ }$’ is initial angular velocity, ‘${\theta _ \circ }$’ is initial angular displacement.‘$\omega $’ is angular velocity, ‘$\alpha $’ is angular acceleration, and ‘$\theta $’ is angular displacement.