**Speed of the Plane:**

Given the position vector function for the plane:

$$r(t) = 10\cos(10\pi t) \boldsymbol{i} + 10\cos(10\pi t) \boldsymbol{j} + (4 + 4t) \boldsymbol{k}, \quad 0 \le t \le \frac{1}{20}.$$

To determine the speed of the plane, we first need to find the velocity vector by taking the derivative of the position vector with respect to time t:

$$v(t) = \frac{dr(t)}{dt}.$$

Differentiating each component of the position vector, we get:

$$v(t) = -100\pi\sin(10\pi t)i - 100\pi\sin(10\pi t)j + 4k.$$

The speed of the plane is the magnitude of the velocity vector:

$$\begin{aligned}\|v(t)\| &=\left\|\frac{dr(t)}{dt}\right\|= \sqrt{(-100\pi\sin(10\pi t))^2 + (-100\pi\sin(10\pi t))^2 + 4^2}\\
&= \sqrt{10000\pi^2\sin^2(10\pi t) + 10000\pi^2\sin^2(10\pi t) + 16}\\
&= \sqrt{20000\pi^2\sin^2(10\pi t) + 16}.\\
\end{aligned}$$

This is the speed of the plane as a function of time.

Tangential and Normal Components of Acceleration

To find the acceleration vector, we need to differentiate the velocity vector with respect to time:

$$a(t) = \frac{dv(t)}{dt}.$$

Differentiating each component of the velocity vector, we get:

$$a(t) = -1000\pi^2\cos(10\pi t)i - 1000\pi^2\cos(10\pi t)j.$$

The tangential component of acceleration ($a_T$) is the projection of the acceleration vector onto the velocity vector. We can find this by taking the dot product of the acceleration and velocity vectors and dividing by the magnitude of the velocity vector [2][4]:

$$a_T = \frac{a(t) \cdot v(t)}{\|v(t)\|}.$$

$$a_T = \frac{-1000\pi^2\cos(10\pi t)(-100\pi\sin(10\pi t)) - 1000\pi^2\cos(10\pi t)(-100\pi\sin(10\pi t))}{\sqrt{20000\pi^2\sin^2(10\pi t) + 16}}$$

$$a_T = \frac{200000\pi^3\sin(10\pi t)\cos(10\pi t)}{\sqrt{20000\pi^2\sin^2(10\pi t) + 16}}$$

The normal component of acceleration ($a_N$) is the component of the acceleration vector perpendicular to the velocity vector [2]. We can find this by taking the cross product of the acceleration and velocity vectors and dividing by the magnitude of the velocity vector:

$$a_N = \frac{\|a(t) \times v(t)\|}{\|v(t)\|}.$$

We have that:
\begin{aligned}
a(t)\times v(t)& \begin{vmatrix} i & j & k \\ -1000\pi^2\cos(10\pi t) & -1000\pi^2\cos(10\pi t) & 0 \\ -100\pi\sin(10\pi t) & -100\pi\sin(10\pi t) & 4 \end{vmatrix}\\
&=i(-4000\pi^2\cos(10\pi t))-j(-4000\pi^2\cos(10\pi t))+k\left(100,0000\pi^3\cos(10\pi t)\sin(10\pi t)-100,0000\pi^3\cos(10\pi t)\sin(10\pi t)\right)\\
&=-i(4000\pi^2\cos(10\pi t))+j(4000\pi^2\cos(10\pi t)).
\end{aligned}

Thus, we have:

\begin{aligned}a_N &= \frac{\sqrt{(4000\pi^2\cos(10\pi t))^2+(4000\pi^2\cos(10\pi t))^2}}{\sqrt{20000\pi^2\sin^2(10\pi t) + 16}}\\
&=\frac{\sqrt{2(4000\pi^2\cos(10\pi t))^2}}{\sqrt{20000\pi^2\sin^2(10\pi t) + 16}}.
\end{aligned}

So, the tangential and normal components of acceleration are:

$$a_T = \frac{200000\pi^3\sin(10\pi t)\cos(10\pi t)}{\sqrt{20000\pi^2\sin^2(10\pi t) + 16}}$$

$$a_N = \frac{\sqrt{2(4000\pi^2\cos(10\pi t))^2}}{\sqrt{20000\pi^2\sin^2(10\pi t) + 16}}.$$