Linear variables in circular motion The following is a really basic problem. I am not interested in the solution rather why the particular solution mentioned below works in all general cases:
Let's imagine a person is running through a circular track. His initial linear speed $u$. The tangential acceleration is $a_t$ and radius of the circular path is $r$. We are asked to find out the time taken to complete one rotation.
The conventional way to do this is $2\pi r=ut+\frac{1}{2}a_{t}t^2$ as consequence of the general equation $s=ut+\frac{1}{2}at^2$. Here are my doubts regarding the equation which I never got to understand while using linear variables for circular motion.
The equation $s=ut+\frac{1}{2}at^2$ was deduced assuming $s$ represents displacement, whereas when we used $s=2\pi r$ in the problem, we didn't take the displacement but rather the distance. Also, there shouldn't be any such term called uniform velocity in circular motion since the direction of velocity vector is constantly changing. But still, I see this term is being used in books. Also the same displacement issue occurs in case of angular displacement. When asked about one or two rotations, we take $\theta$ to be a multiple of $2\pi$ radians whereas the displacement is $0$ radians.
I am a novice in physics so I would like to have the kind attention of the respected users here to rectify my misconceptions.
 A: The original problem statement gave you a linear speed, not a linear velocity, because as you surmised, velocity has both a speed and a direction.  For circular motion, the speed may or not be constant, but it is certain that the direction of the motion is continuously changing, so declaring a velocity is technically invalid.  Also note that the problem statement asked for time of travel, not displacement, so it is OK to use distance traveled for this problem.
Regarding what the problem is asking for, the equation $2\pi r=ut+\frac{1}{2}a_{t}t^2$ does indeed allow you to calculate the time that it takes to make one complete revolution, because the tangential acceleration is constant (a requirement of the kinematic equations), and that equation will be solved with the quadratic formula.  A bit of care will be required when arriving at a solution because only one of the roots of the quadratic formula will be what you are looking for.
A: R.W. Bird is correct that there's some confusion re. terminology. However, as usual, meaning can be derived from context.
Now look  at the situation above:
The runner runs from $A$ to $B$, following the arc. The  angular displacement here is $\theta$.
The angular velocity $\omega$ is given by:
$$\omega=\frac{\mathrm{d}\theta}{\mathrm{d}t}$$
If the motion is not uniform, then there must be angular acceleration $\alpha$:
$$\alpha=\frac{\mathrm{d}\omega}{\mathrm{d}t}=\frac{\mathrm{d^2}\theta}{\mathrm{d}t^2}$$
In the case of constant angular acceleration $\alpha$ the angular displacement $\theta(t)$ is given by:
$$\theta(t)=\theta_0+\omega_0t+\frac12 \alpha t^2$$
where the suffix $_0$ refers to values at $t=0$.
The angular and linear displacements relate as follows:
$$s(t)=\theta(t)R$$
with $R$ the radius of the circular track.

How do we derive  $\theta(t)=\theta_0+\omega_0t+\frac12 \alpha t^2$?
Firstly we have, with $\omega(0)=\omega_0$, so that:
$$\alpha=\frac{\mathrm{d}\omega}{\mathrm{d}t}$$
$$\int_{\omega_0}^{\omega(t)}\mathrm{d}\omega=\alpha\int_0^t\mathrm{d}t$$
$$\Rightarrow \omega(t)=\omega_0+\alpha t$$
Then with $\omega=\frac{\mathrm{d}\theta}{\mathrm{d}t}$ and $\theta(0)=\theta_0$:
$$\int_{\theta_0}^{\theta(t)}\mathrm{d}\theta=\int_0^t \omega(t) \mathrm{d}t=\int_0^t\left(\omega_0+\alpha t\right)\mathrm{d}t$$
$$\boxed{\theta(t)=\theta_0+\omega_0t+\frac12 \alpha t^2}$$
A: This is an unusual problem and is therefore a bit confusing.  Usually a runner on a circular track moves at a constant speed and is subject to a centripetal acceleration. In this case, you are given a tangential acceleration. The runner's (tangential) speed is increasing and the formula you have for the distance around the track (an arc length) is correct.  You can solve it for the time.  Technically, the term “displacement” refers to a vector.  After completing one circuit of the track, the runner's displacement is zero.
