A wheel rotates around a stationary axis so that the rotation angle $\varphi =a{t}^2$, where $a=0.20 rad/s^2$. Find the total acceleration $a$ of the point $A$ at the rim at the moment $t=2.5 s$ if the linear velocity of the point $A$ at this moment $v=0.65 m/s$.

Given, $\varphi =a{t}^2$

Angular velocity, $\omega =\frac{d\varphi }{dt}=2at$

$\alpha =\frac{d^2\varphi }{dt}=2a$

we know that,

 $$\begin{gathered} v=\omega r \\ v=\left(2at\right)r \\ \Rightarrow r=\frac{v}{2at} \\ =\frac{0.65}{2\times 0.2\times 2.5}=0.65m \end{gathered}$$

Tangential acceleration 

$$\begin{gathered} a_t=\alpha r \\ =2ar \\ =2\times 0.20\times 0.65 \\ =0.26 m/s^2 \end{gathered}$$

Normal acceleration

$$\begin{gathered} a_c=\frac{v^2}{r} \\ =\frac{0.{65}^2}{0.65} \\ =0.65 m/s^2 \end{gathered}$$

Total acceleration,

$a=\sqrt{{a_t}^2+{a_c}^2}=0.7 m/s^2$