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# System prepended metadata

title: PID-826

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To answer the questions, we first need to find the velocity and acceleration vectors of the plane [1]. The position vector of the plane is given by:

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

**Velocity and Speed:**

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

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

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:

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

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