The first step is to decide where to put the origin of the rotating coordinates system, and the second step is to decide to where to attach it to.
Lets put the origin at point \(B\) and have the frame attached to the bar \(BD\) as well. This way the relative velocity and acceleration will be simple, but the angular acceleration will be more involved.
Therefore, this diagram shows a general configuration to help understand the set up
Let units vectors for rotating coordinates system be \(\overset{\rightarrow }{i},\overset{\rightarrow }{j},\overset{\rightarrow }{k}\) and for the fixed coordinates system be \(\overset{\rightarrow }{I},\overset{\rightarrow }{J},\overset{\rightarrow }{K}\).
From the above, Let \(L\) be the length of Bar AB. Hence\begin{align*} \overset{\rightarrow }{\rho } & =\rho \overset{\rightarrow }{j}\\ \dot{\overset{\rightarrow }{\rho }}_{r} & =\dot{\rho }\overset{\rightarrow }{j}\\ \dot{\overset{\rightarrow }{R}} & =L\omega _{1}\overset{\rightarrow }{j}\qquad \text{at snapshot only}\\ \overset{\rightarrow }{\omega } & =\omega _{1}\overset{\rightarrow }{I}+\omega _{2}\overset{\rightarrow }{k}\qquad \end{align*}
Hence \begin{align} \overset{\rightarrow }{V} & =\dot{\overset{\rightarrow }{R}}+\dot{\overset{\rightarrow }{\rho }}_{r}+\overset{\rightarrow }{\omega }\times \overset{\rightarrow }{\rho }\tag{1}\\ & =L\omega _{1}\overset{\rightarrow }{j}+\dot{\rho }\overset{\rightarrow }{j}+\left ( \omega _{1}\overset{\rightarrow }{I}+\omega _{2}\overset{\rightarrow }{k}\right ) \times \rho \overset{\rightarrow }{j}\nonumber \end{align}
But at snapshot, \(\overset{\rightarrow }{I}=\overset{\rightarrow }{i}\), hence
\begin{align*} \overset{\rightarrow }{V} & =L\omega _{1}\overset{\rightarrow }{j}+\dot{\rho }\overset{\rightarrow }{j}+\omega _{1}\rho \overset{\rightarrow }{k}-\omega _{2}\rho \overset{\rightarrow }{i}\\ & =-\omega _{2}\rho \overset{\rightarrow }{i}+\left ( L\omega _{1}+\dot{\rho }\right ) \overset{\rightarrow }{j}+\omega _{1}\rho \overset{\rightarrow }{k} \end{align*}
At snapshot, \(\omega _{2}=5rad/\sec ,\omega _{1}=4rad/\sec ,L=0.5m,\dot{\rho }=3m/s,\rho =0.2m\), hence\begin{align} \overset{\rightarrow }{V} & =-5\left ( 0.2\right ) \overset{\rightarrow }{i}+\left ( 0.5\left ( 4\right ) +3\right ) \overset{\rightarrow }{j}+4\left ( 0.2\right ) \overset{\rightarrow }{k}\nonumber \\ & =-1\overset{\rightarrow }{i}+5\overset{\rightarrow }{j}+0.8\overset{\rightarrow }{k} \tag{2} \end{align}
Hence \[ \left \vert \overset{\rightarrow }{V}\right \vert =\sqrt{1^{2}+5^{2}+0.8^{2}}=5.1614\qquad m/sec \]
Now to find the acceleration\begin{align} \overset{\rightarrow }{a} & =\ddot{\overset{\rightarrow }{R}}+\overset{\centerdot \centerdot }{\overset{\rightarrow }{\rho }}_{r}+\overset{\rightarrow }{\omega }\times \dot{\overset{\rightarrow }{\rho }}_{r}+\dot{\overset{\rightarrow }{\omega }}\times \overset{\rightarrow }{\rho }+\overset{\rightarrow }{\omega }\times \left ( \dot{\overset{\rightarrow }{\rho }}_{r}+\overset{\rightarrow }{\omega }\times \overset{\rightarrow }{\rho }\right ) \nonumber \\ & =\ddot{\overset{\rightarrow }{R}}+\ddot{\overset{\rightarrow }{\rho }}_{r}+\overset{\rightarrow }{\omega }\times \dot{\overset{\rightarrow }{\rho }}_{r}+\dot{\overset{\rightarrow }{\omega }}\times \overset{\rightarrow }{\rho }+\overset{\rightarrow }{\omega }\times \dot{\overset{\rightarrow }{\rho }}_{r}+\left ( \overset{\rightarrow }{\omega }\times \left ( \overset{\rightarrow }{\omega }\times \overset{\rightarrow }{\rho }\right ) \right ) \nonumber \\ & =\ddot{\overset{\rightarrow }{R}}+\ddot{\overset{\rightarrow }{\rho }}_{r}+2\overset{\rightarrow }{\omega }\times \dot{\overset{\rightarrow }{\rho }}_{r}+\dot{\overset{\rightarrow }{\omega }}\times \overset{\rightarrow }{\rho }+\left ( \overset{\rightarrow }{\omega }\times \left ( \overset{\rightarrow }{\omega }\times \overset{\rightarrow }{\rho }\right ) \right ) \tag{3} \end{align}
Now each term is found. \begin{align*} \ddot{\overset{\rightarrow }{R}} & =L\dot{\omega }_{1}\overset{\rightarrow }{j}+L\omega _{1}^{2}\overset{\rightarrow }{k}=0.5\left ( 1.5\right ) \overset{\rightarrow }{j}+0.5\left ( 4^{2}\right ) \overset{\rightarrow }{k}=0.75\overset{\rightarrow }{j}+8\overset{\rightarrow }{k}\\ \ddot{\overset{\rightarrow }{\rho }}_{r} & =\ddot{\rho }\overset{\rightarrow }{j}\\ \dot{\overset{\rightarrow }{\omega }} & =\overset{\centerdot }{\omega }_{1}\overset{\rightarrow }{I}-\overset{\frac{d}{dt}\left ( \omega _{2}\overset{\rightarrow }{k}\right ) }{\overbrace{\left ( \dot{\omega }_{2}\overset{\rightarrow }{k}+\left ( \omega _{1}\overset{\rightarrow }{i}\times \omega _{2}\overset{\rightarrow }{k}\right ) \right ) }}\\ \dot{\overset{\rightarrow }{\omega }} & =\dot{\omega }_{1}\overset{\rightarrow }{I}-\dot{\omega }_{2}\overset{\rightarrow }{k}+\omega _{1}\omega _{2}\overset{\rightarrow }{j} \end{align*}
But at snapshot \(\overset{\rightarrow }{I}=\overset{\rightarrow }{i}\), hence
\[ \dot{\overset{\rightarrow }{\omega }}=\overset{\centerdot }{\omega }_{1}\overset{\rightarrow }{i}-\dot{\omega }_{2}\overset{\rightarrow }{k}+\omega _{1}\omega _{2}\overset{\rightarrow }{j}\]
Now all the terms have been found, then Eq. (3) becomes (valid at snapshot only)\begin{align*} \overset{\rightarrow }{a} & =\left ( L\dot{\omega }_{1}\overset{\rightarrow }{j}+L\omega _{1}^{2}\overset{\rightarrow }{k}\right ) +\ddot{\rho }\overset{\rightarrow }{j}\\ & +2\left ( \omega _{1}\overset{\rightarrow }{i}+\omega _{2}\overset{\rightarrow }{k}\right ) \times \dot{\rho }\overset{\rightarrow }{j}\\ & +\left ( \dot{\omega }_{1}\overset{\rightarrow }{i}-\dot{\omega }_{2}\overset{\rightarrow }{k}+\omega _{1}\omega _{2}\overset{\rightarrow }{j}\right ) \times \rho \overset{\rightarrow }{j}\\ & +\left ( \left ( \omega _{1}\overset{\rightarrow }{i}+\omega _{2}\overset{\rightarrow }{k}\right ) \times \left ( \left ( \omega _{1}\overset{\rightarrow }{i}+\omega _{2}\overset{\rightarrow }{k}\right ) \times \rho \overset{\rightarrow }{j}\right ) \right ) \end{align*}
Hence\begin{align*} \overset{\rightarrow }{a} & =\left ( L\dot{\omega }_{1}\overset{\rightarrow }{j}+L\omega _{1}^{2}\overset{\rightarrow }{k}\right ) +\ddot{\rho }\overset{\rightarrow }{j}\\ & +2\left ( \omega _{1}\dot{\rho }\overset{\rightarrow }{k}-\omega _{2}\dot{\rho }\overset{\rightarrow }{i}\right ) \\ & +\left ( \dot{\omega }_{1}\rho \overset{\rightarrow }{k}+\dot{\omega }_{2}\rho \overset{\rightarrow }{i}\right ) \\ & +\left ( \left ( \omega _{1}\overset{\rightarrow }{i}+\omega _{2}\overset{\rightarrow }{k}\right ) \times \left ( \omega _{1}\rho \overset{\rightarrow }{k}-\omega _{2}\rho \overset{\rightarrow }{i}\right ) \right ) \end{align*}
Therefore\[ \overset{\rightarrow }{a}=\left ( L\dot{\omega }_{1}\overset{\rightarrow }{j}+L\omega _{1}^{2}\overset{\rightarrow }{k}\right ) +\ddot{\rho }\overset{\rightarrow }{j}+2\left ( \omega _{1}\dot{\rho }\overset{\rightarrow }{k}-\omega _{2}\dot{\rho }\overset{\rightarrow }{i}\right ) +\left ( \dot{\omega }_{1}\rho \overset{\rightarrow }{k}+\dot{\omega }_{2}\rho \overset{\rightarrow }{i}\right ) +\left ( -\omega _{1}\omega _{1}\rho \overset{\rightarrow }{j}-\omega _{2}\omega _{2}\rho \overset{\rightarrow }{j}\right ) \]
Collecting terms\[ \overset{\rightarrow }{a}=\left ( -2\omega _{2}\dot{\rho }+\dot{\omega }_{2}\rho \right ) \overset{\rightarrow }{i}+\left ( L\dot{\omega }_{1}+\overset{\centerdot \centerdot }{\rho }-\omega _{1}^{2}\rho -\omega _{2}^{2}\rho \right ) \overset{\rightarrow }{j}+\left ( L\omega _{1}^{2}+2\omega _{1}\dot{\rho }+\dot{\omega }_{1}\rho \right ) \overset{\rightarrow }{k}\]
At snapshot, \(\omega _{2}=5rad/\sec ,\omega _{1}=4rad/\sec ,L=0.5m,\dot{\rho }=3m/s,\rho =0.2m,\overset{\centerdot \centerdot }{\rho }=2m/s,\dot{\omega }_{1}=1.5rad/\sec ^{2},\dot{\omega }_{2}=6m/\sec ^{2}\), hence\begin{align*} \overset{\rightarrow }{a} & =\left ( -\left ( 2\right ) 5\left ( 3\right ) +6\left ( 0.2\right ) \right ) \overset{\rightarrow }{i}+\left ( 0.5\left ( 1.5\right ) +2-4^{2}\left ( 0.2\right ) -5^{2}\left ( 0.2\right ) \right ) \overset{\rightarrow }{j}+\left ( 0.5\left ( 4^{2}\right ) +2\left ( 4\right ) 3+1.5\left ( 0.2\right ) \right ) \overset{\rightarrow }{k}\\ & =-28.8\overset{\rightarrow }{i}-5.45\overset{\rightarrow }{j}+32.3\overset{\rightarrow }{k} \end{align*}
Hence\[ \left \vert \overset{\rightarrow }{a}\right \vert =\sqrt{28.8^{2}+5.45^{2}+32.3^{2}}=43.617m/s^{2}\]
Let the origin of the rotating frame be \(G\) as shown. Let \(L\) be the length given by \(25^{\prime }\), and let \(r\) be the radius of the hydraulic line. Hence, for the fluid particle at B\begin{align*} \overset{\rightarrow }{\rho } & =\left ( L+r\right ) \overset{\rightarrow }{j}+r\overset{\rightarrow }{k}\\ \dot{\overset{\rightarrow }{\rho }}_{r} & =\overset{\centerdot }{s}\overset{\rightarrow }{k}\\ \dot{\overset{\rightarrow }{R}} & =200\overset{\rightarrow }{j}+3t\overset{\rightarrow }{k}\\ \overset{\rightarrow }{\omega } & =\dot{\alpha }\overset{\rightarrow }{i}+\dot{\beta }\overset{\rightarrow }{j} \end{align*}
Hence \begin{align} \overset{\rightarrow }{V} & =\dot{\overset{\rightarrow }{R}}+\dot{\overset{\rightarrow }{\rho }}_{r}+\overset{\rightarrow }{\omega }\times \overset{\rightarrow }{\rho }\tag{1}\\ & =\left ( 200\overset{\rightarrow }{j}+3t\overset{\rightarrow }{k}\right ) +\left ( \dot{s}\overset{\rightarrow }{k}\right ) +\left ( \dot{\alpha }\overset{\rightarrow }{i}+\overset{\centerdot }{\beta }\overset{\rightarrow }{j}\right ) \times \left ( \left ( L+r\right ) \overset{\rightarrow }{j}+r\overset{\rightarrow }{k}\right ) \nonumber \\ & =\left ( 200\overset{\rightarrow }{j}+3t\overset{\rightarrow }{k}\right ) +\left ( \dot{s}\overset{\rightarrow }{k}\right ) +\dot{\alpha }\left ( L+r\right ) \overset{\rightarrow }{k}-\dot{\alpha }r\overset{\rightarrow }{j}+\overset{\centerdot }{\beta }r\overset{\rightarrow }{i}\nonumber \\ & =\overset{\rightarrow }{i}\left ( \dot{\beta }r\right ) +\overset{\rightarrow }{j}\left ( 200-\dot{\alpha }r\right ) +\overset{\rightarrow }{k}\left ( 3t+\dot{s}+\dot{\alpha }\left ( L+r\right ) \right ) \nonumber \end{align}
At snapshot, \(t=10\sec ,r=5^{\prime },L=25^{\prime },\dot{\beta }=0.2rad/\sec ,\dot{\alpha }=0.1rad/\sec ,\overset{\centerdot }{s}=70-5t\), hence the above becomes\begin{align*} \overset{\rightarrow }{V} & =\overset{\rightarrow }{i}\left ( 0.2\left ( 5\right ) \right ) +\overset{\rightarrow }{j}\left ( 200-0.1\left ( 5\right ) \right ) +\overset{\rightarrow }{k}\left ( 3\left ( 10\right ) +\left ( 70-50\right ) +0.1\left ( 25+5\right ) \right ) \\ & =\overset{\rightarrow }{i}+199.\,5\overset{\rightarrow }{j}+53\overset{\rightarrow }{k} \end{align*}
Hence \[ \left \vert \overset{\rightarrow }{V}\right \vert =\sqrt{1^{2}+199.5^{2}+53^{2}}=206.42\text{ ft/sec}\]
To find the acceleration\begin{align} \overset{\rightarrow }{a} & =\overset{\centerdot \centerdot }{\overset{\rightarrow }{R}}+\overset{\centerdot \centerdot }{\overset{\rightarrow }{\rho }}_{r}+\overset{\rightarrow }{\omega }\times \dot{\overset{\rightarrow }{\rho }}_{r}+\dot{\overset{\rightarrow }{\omega }}\times \overset{\rightarrow }{\rho }+\overset{\rightarrow }{\omega }\times \left ( \overset{\centerdot }{\overset{\rightarrow }{\rho }}_{r}+\overset{\rightarrow }{\omega }\times \overset{\rightarrow }{\rho }\right ) \nonumber \\ & =\ddot{\overset{\rightarrow }{R}}+\ddot{\overset{\rightarrow }{\rho }}_{r}+\overset{\rightarrow }{\omega }\times \dot{\overset{\rightarrow }{\rho }}_{r}+\dot{\overset{\rightarrow }{\omega }}\times \overset{\rightarrow }{\rho }+\overset{\rightarrow }{\omega }\times \dot{\overset{\rightarrow }{\rho }}_{r}+\left ( \overset{\rightarrow }{\omega }\times \left ( \overset{\rightarrow }{\omega }\times \overset{\rightarrow }{\rho }\right ) \right ) \nonumber \\ & =\ddot{\overset{\rightarrow }{R}}+\ddot{\overset{\rightarrow }{\rho }}_{r}+2\overset{\rightarrow }{\omega }\times \dot{\overset{\rightarrow }{\rho }}_{r}+\dot{\overset{\rightarrow }{\omega }}\times \overset{\rightarrow }{\rho }+\left ( \overset{\rightarrow }{\omega }\times \left ( \overset{\rightarrow }{\omega }\times \overset{\rightarrow }{\rho }\right ) \right ) \tag{2} \end{align}
Now each term is found. \begin{align*} \ddot{\overset{\rightarrow }{R}} & =\frac{d}{dt}\left ( \dot{\overset{\rightarrow }{R}}\right ) =3\overset{\rightarrow }{k}+\left ( \dot{\alpha }\overset{\rightarrow }{i}+\dot{\beta }\overset{\rightarrow }{j}\right ) \times \left ( 200\overset{\rightarrow }{j}+3t\overset{\rightarrow }{k}\right ) \\ & =3\overset{\rightarrow }{k}+\left ( 0.1\overset{\rightarrow }{i}+0.2\overset{\rightarrow }{j}\right ) \times \left ( 200\overset{\rightarrow }{j}+3t\overset{\rightarrow }{k}\right ) \\ & =3\overset{\rightarrow }{k}+20\overset{\rightarrow }{k}-0.3t\overset{\rightarrow }{j}+0.6t\overset{\rightarrow }{i}\\ & =0.6t\overset{\rightarrow }{i}-0.3t\overset{\rightarrow }{j}+23\overset{\rightarrow }{k}\\ \ddot{\overset{\rightarrow }{\rho }}_{r} & =\ddot{s}\overset{\rightarrow }{k}-\frac{\dot{s}^{2}}{r}\overset{\rightarrow }{j}\\ \dot{\overset{\rightarrow }{\omega }} & =\overset{\frac{d}{dt}\left ( \dot{\alpha }\overset{\rightarrow }{i}\right ) }{\overbrace{\ddot{\alpha }\overset{\rightarrow }{i}+\left ( \dot{\beta }\overset{\rightarrow }{j}\times \dot{\alpha }\overset{\rightarrow }{i}\right ) }}+\overset{\frac{d}{dt}\left ( \dot{\beta }\overset{\rightarrow }{j}\right ) }{\overbrace{\left ( \ddot{\beta }\overset{\rightarrow }{j}+\left ( \dot{\alpha }\overset{\rightarrow }{i}\times \dot{\beta }\overset{\rightarrow }{j}\right ) \right ) }}\\ \dot{\overset{\rightarrow }{\omega }} & =\overset{\centerdot \centerdot }{\alpha }\overset{\rightarrow }{i}-\dot{\beta }\dot{\alpha }\overset{\rightarrow }{k}+\overset{\centerdot \centerdot }{\beta }\overset{\rightarrow }{j}+\dot{\alpha }\dot{\beta }\overset{\rightarrow }{k}\\ & =\ddot{\alpha }\overset{\rightarrow }{i}+\ddot{\beta }\overset{\rightarrow }{j} \end{align*}
Now all the terms have been found, then Eq. (2) becomes\begin{align*} \overset{\rightarrow }{a} & =\overset{\centerdot \centerdot }{\overset{\rightarrow }{R}}+\overset{\centerdot \centerdot }{\overset{\rightarrow }{\rho }}_{r}+2\overset{\rightarrow }{\omega }\times \dot{\overset{\rightarrow }{\rho }}_{r}+\dot{\overset{\rightarrow }{\omega }}\times \overset{\rightarrow }{\rho }+\left ( \overset{\rightarrow }{\omega }\times \left ( \overset{\rightarrow }{\omega }\times \overset{\rightarrow }{\rho }\right ) \right ) \\ & =\left ( 0.6t\overset{\rightarrow }{i}-0.3t\overset{\rightarrow }{j}+23\overset{\rightarrow }{k}\right ) +\overset{\centerdot \centerdot }{s}\overset{\rightarrow }{k}-\frac{\dot{s}^{2}}{r}\overset{\rightarrow }{j}\\ & +2\left ( \dot{\alpha }\overset{\rightarrow }{i}+\dot{\beta }\overset{\rightarrow }{j}\right ) \times \dot{s}\overset{\rightarrow }{k}\\ & +\left ( \ddot{\alpha }\overset{\rightarrow }{i}+\ddot{\beta }\overset{\rightarrow }{j}\right ) \times \left ( \left ( L+r\right ) \overset{\rightarrow }{j}+r\overset{\rightarrow }{k}\right ) \\ & +\left ( \dot{\alpha }\overset{\rightarrow }{i}+\dot{\beta }\overset{\rightarrow }{j}\right ) \times \left ( \left ( \dot{\alpha }\overset{\rightarrow }{i}+\dot{\beta }\overset{\rightarrow }{j}\right ) \times \left ( \left ( L+r\right ) \overset{\rightarrow }{j}+r\overset{\rightarrow }{k}\right ) \right ) \end{align*}
Hence\begin{align*} \overset{\rightarrow }{a} & =\left ( 0.6t\overset{\rightarrow }{i}-0.3t\overset{\rightarrow }{j}+23\overset{\rightarrow }{k}\right ) +\ddot{s}\overset{\rightarrow }{k}-\frac{\dot{s}^{2}}{r}\overset{\rightarrow }{j}\\ & +2\left ( -\dot{\alpha }\dot{s}\overset{\rightarrow }{j}+\dot{\beta }\overset{\centerdot }{s}\overset{\rightarrow }{i}\right ) \\ & +\ddot{\alpha }\left ( L+r\right ) \overset{\rightarrow }{k}-\ddot{\alpha }r\overset{\rightarrow }{j}+\ddot{\beta }r\overset{\rightarrow }{i}\\ & +\left ( \dot{\alpha }\overset{\rightarrow }{i}+\dot{\beta }\overset{\rightarrow }{j}\right ) \times \left ( \dot{\alpha }\left ( L+r\right ) \overset{\rightarrow }{k}-\dot{\alpha }r\overset{\rightarrow }{j}+\overset{\centerdot }{\beta }r\overset{\rightarrow }{i}\right ) \end{align*}
Therefore\begin{align*} \overset{\rightarrow }{a} & =\left ( 0.6t\overset{\rightarrow }{i}-0.3t\overset{\rightarrow }{j}+23\overset{\rightarrow }{k}\right ) +\ddot{s}\overset{\rightarrow }{k}-\frac{\dot{s}^{2}}{r}\overset{\rightarrow }{j}\\ & +2\left ( -\dot{\alpha }\dot{s}\overset{\rightarrow }{j}+\dot{\beta }\overset{\centerdot }{s}\overset{\rightarrow }{i}\right ) \\ & +\ddot{\alpha }\left ( L+r\right ) \overset{\rightarrow }{k}-\ddot{\alpha }r\overset{\rightarrow }{j}+\ddot{\beta }r\overset{\rightarrow }{i}\\ & -\dot{\alpha }^{2}\left ( L+r\right ) \overset{\rightarrow }{j}-\dot{\alpha }^{2}r\overset{\rightarrow }{k}+\dot{\beta }\dot{\alpha }\left ( L+r\right ) \overset{\rightarrow }{i}-\dot{\beta }^{2}r\overset{\rightarrow }{k} \end{align*}
Collecting terms\[ \overset{\rightarrow }{a}=\overset{\rightarrow }{i}\left ( 0.6t+2\dot{\beta }\dot{s}+\overset{\centerdot \centerdot }{\beta }r+\dot{\beta }\dot{\alpha }\left ( L+r\right ) \right ) +\overset{\rightarrow }{j}\left ( -0.3t-\frac{\dot{s}^{2}}{r}-2\dot{\alpha }\dot{s}-\ddot{\alpha }r-\dot{\alpha }^{2}\left ( L+r\right ) \right ) +\overset{\rightarrow }{k}\left ( 23+\overset{\centerdot \centerdot }{s}+\ddot{\alpha }\left ( L+r\right ) -\dot{\alpha }^{2}r-\dot{\beta }^{2}r\right ) \]
Since angular accelerations are constants, the above simplifies to\[ \overset{\rightarrow }{a}=\overset{\rightarrow }{i}\left ( 0.6t+2\dot{\beta }\dot{s}+\overset{\centerdot }{\beta }\dot{\alpha }\left ( L+r\right ) \right ) +\overset{\rightarrow }{j}\left ( -0.3t-\frac{\dot{s}^{2}}{r}-2\dot{\alpha }\dot{s}-\overset{\centerdot }{\alpha }^{2}\left ( L+r\right ) \right ) +\overset{\rightarrow }{k}\left ( 23+\ddot{s}-\dot{\alpha }^{2}r-\dot{\beta }^{2}r\right ) \]
Now \(\ddot{s}=-5,\) hence at snapshot where, \(t=10\sec ,r=5^{\prime },L=25^{\prime },\dot{\beta }=0.2rad/\sec ,\dot{\alpha }=0.1rad/\sec ,\overset{\centerdot }{s}=70-5t\) the above becomes\begin{align*} \overset{\rightarrow }{a} & =\overset{\rightarrow }{i}\left ( 6+2\left ( 0.2\right ) \left ( 70-50\right ) +0.2\left ( 0.1\right ) \left ( 25+5\right ) \right ) +\overset{\rightarrow }{j}\left ( -3-\frac{\left ( 70-50\right ) ^{2}}{5}-2\left ( 0.1\right ) \left ( 70-50\right ) -0.1^{2}\left ( 25+5\right ) \right ) \\ & +\overset{\rightarrow }{k}\left ( 23-5-0.1^{2}\left ( 5\right ) -0.2^{2}\left ( 5\right ) \right ) \\ & \\ & =14.6\overset{\rightarrow }{i}-87.3\overset{\rightarrow }{j}+17.75\overset{\rightarrow }{k} \end{align*}
Hence\[ \left \vert \overset{\rightarrow }{a}\right \vert =\sqrt{14.6^{2}+87.3^{2}+17.75^{2}}=90.275\text{ ft/sec}^{2}\]