Engine crank 1800 rpm, connecting rod velocity

Instantaneous centre of rotation (ICR) is the point in the plane of motion (on the body or on its imaginary extension) whose velocity is zero at that instant. At that instant the body behaves as if it were rotating about this point, so every point has speed $v = \omega\,r$, where $r$ is its distance from the ICR, and its velocity is perpendicular to the line joining it to the ICR.

Geometry

Take $A$ as origin, $x$ horizontal towards the piston, $y$ upward. The crank turns clockwise, so $\omega_{AB} = -188.50$ rad/s ($1800$ rpm $= 1800\times 2\pi/60$). The piston $D$ moves only horizontally, so $y_D = 0$:

$$\sin\beta = \frac{AB\sin 40^\circ}{BD} = \frac{3\sin 40^\circ}{8} = 0.2410 \;\Rightarrow\; \beta = 13.95^\circ$$ $$\vec r_{B/A} = (2.2981\,\hat i + 1.9284\,\hat j)\ \text{in}, \qquad \vec r_{D/B} = (7.7641\,\hat i - 1.9284\,\hat j)\ \text{in}$$

Velocities

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

$$\vec v_B = \vec\omega_{AB}\times\vec r_{B/A} = (363.49\,\hat i -433.19\,\hat j)\ \text{in/s}, \qquad v_B = \omega_{AB}\,AB = 565.49\ \text{in/s}$$

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

$$\begin{aligned} \hat j:&\quad 0 = -433.19 + \omega_{BD}(7.7641) \;\Rightarrow\; \omega_{BD} = 55.79\ \text{rad/s}\\ \hat i:&\quad v_D = 363.49 + \omega_{BD}(1.9284) = 471.08\ \text{in/s} \end{aligned}$$

The positive $\omega_{BD}$ means the rod turns counter-clockwise.

Answer: (a) $\omega_{BD} \approx 55.79$ rad/s counter-clockwise; (b) $v_P = v_D \approx 471.08$ in/s $= 39.26$ ft/s (to the right).