A train ride

We have a uniform circular motion because the trajectory is a circumference and the speed does not change throughout the motion.

Question a)

Data

R = 4 m
T = 10 s

Resolution

To calculate the angular velocity, we will use the following expression:

$$\begin{gathered} \omega =\frac{2·\pi }{T}\Rightarrow \\ \omega =\frac{6.28 rad}{10 s}\Rightarrow \\ \boxed{\omega =0.628 \raisebox{1ex}{$rad$}\!\left/ \!\raisebox{-1ex}{$s$}\right.} \end{gathered}$$

And the linear velocity is:

$$\begin{gathered} v=\omega ·R\Rightarrow \\ v=0.628 \raisebox{1ex}{$rad$}\!\left/ \!\raisebox{-1ex}{$s$}\right.·2 m\Rightarrow \\ \boxed{v=1.26 \raisebox{1ex}{$m$}\!\left/ \!\raisebox{-1ex}{$s$}\right.} \end{gathered}$$

Question b)

Data

ω = 0.628 m
t = 2 min = 120 s
R = 2 m
φ₀ = 0 rad (assume that the initial angle is 0 rad).
s₀ = 0 m (assume that the initial distance traveled is 0 m).

Resolution

To calculate the angle covered:

$$\begin{gathered} \phi ={\phi }_0+\omega ·t \Rightarrow \\ \phi =0 rad+0.628 \raisebox{1ex}{$rad$}\!\left/ \!\raisebox{-1ex}{$\cancel{s}$}\right.·120 \cancel{s} \Rightarrow \\ \boxed{\phi = 75.36 rad} \end{gathered}$$

and the distance traveled:

$$s=\phi ·R \Rightarrow \boxed{ s=75.36 rad·2 m =150.72 m}$$

Question c)

Since we are dealing with a u.c.m. the values of the accelerations that we can calculate in this type of motion are:

$${\displaystyle \boxed{\alpha =0 {\displaystyle \raisebox{1ex}{$rad$}\!\left/ \!\raisebox{-1ex}{$s^2$}\right.}} ⋮\boxed{a_n=\frac{v^2}{R}={\omega }^2·R = 0.7887 m/s^2}⋮\boxed{ a_t=0 {\displaystyle \raisebox{1ex}{$m$}\!\left/ \!\raisebox{-1ex}{$s^2$}\right.}}}$$