Engine crank 200 rpm, rod and piston acceleration

Geometry

Take $A$ as origin, $x$ horizontal towards the piston, $y$ upward. The crank turns clockwise, so $\omega_{AB} = -20.94$ rad/s ($200$ rpm $= 200\times 2\pi/60$). The piston $D$ moves only horizontally, so $y_D = 0$:

$$\sin\beta = \frac{AB\sin 60^\circ}{BD} = \frac{0.15\sin 60^\circ}{0.5} = 0.2598 \;\Rightarrow\; \beta = 15.06^\circ$$ $$\vec r_{B/A} = (0.0750\,\hat i + 0.1299\,\hat j)\ \text{m}, \qquad \vec r_{D/B} = (0.4828\,\hat i - 0.1299\,\hat j)\ \text{m}$$

Velocities

Crank pin $B$ moves on a circle about $A$:

$$\vec v_B = \vec\omega_{AB}\times\vec r_{B/A} = (2.72\,\hat i -1.57\,\hat j)\ \text{m/s}, \qquad v_B = \omega_{AB}\,AB = 3.14\ \text{m/s}$$

For the connecting rod, $\vec v_D = \vec v_B + \omega_{BD}\hat k\times\vec r_{D/B}$ with $\vec v_D = v_D\hat i$:

$$\begin{aligned} \hat j:&\quad 0 = -1.57 + \omega_{BD}(0.4828) \;\Rightarrow\; \omega_{BD} = 3.25\ \text{rad/s}\\ \hat i:&\quad v_D = 2.72 + \omega_{BD}(0.1299) = 3.14\ \text{m/s} \end{aligned}$$

Accelerations

The crank speed is constant ($\alpha_{AB}=0$), so $B$ has only a normal acceleration towards $A$:

$$\vec a_B = -\omega_{AB}^2\,\vec r_{B/A} = (-32.9\,\hat i -57.0\,\hat j)\ \text{m/s}^2 \quad (a_B = 65.8\ \text{m/s}^2)$$

For the rod, with $\omega_{BD} = 3.25$ rad/s (counter-clockwise) and $\vec a_D = a_D\hat i$:

$$\vec a_D = \vec a_B + \alpha_{BD}\hat k\times\vec r_{D/B} - \omega_{BD}^2\,\vec r_{D/B}$$ $$\omega_{BD}^2\,\vec r_{D/B} = (5.1\,\hat i -1.4\,\hat j)\ \text{m/s}^2, \qquad \hat k\times\vec r_{D/B} = (0.1299\,\hat i + 0.4828\,\hat j)\ \text{m}$$

Equating components:

$$\begin{aligned} \hat j:&\quad 0 = -57.0 +1.4 + \alpha_{BD}(0.4828) \;\Rightarrow\; \alpha_{BD} = 115.2\ \text{rad/s}^2\\ \hat i:&\quad a_D = -32.9 -5.1 + \alpha_{BD}(0.1299) = -23.0\ \text{m/s}^2 \end{aligned}$$

Answer: $\alpha_{BD} \approx 115.2$ rad/s² (counter-clockwise), $a_D \approx 23.0$ m/s² (towards the crank, i.e. to the left).