## Step 1: Identify the Velocity Direction for Point A - The bar \( O_2A \) is rotating **counterclockwise** about \( O_2 \) with angular velocity \( \omega_2 \). - The **velocity of A (\( \vec{V}_A \))** is always perpendicular to \( O_2A \), and the direction is tangent to the circle that A traces around \( O_2 \), following the right-hand rule for CCW rotation. **Explanation:** For a point on a rotating link, the velocity vector at that point is perpendicular to the link and points in the direction of rotation (CCW in this case). --- ## Step 2: Estimate the Angle Relative to the Positive X-Axis - The link \( O_2A \) is shown at an angle above the x-axis (horizontal). - Since \( \vec{V}_A \) is perpendicular to \( O_2A \) and \( O_2A \) is above the x-axis, the perpendicular direction is also above the x-axis, but rotated 90° **ahead** of the angle of \( O_2A \). - Visually estimating from the diagram, \( O_2A \) is about 37° above the x-axis. So, \( \vec{V}_A \) would be **90° ahead** (CCW) of this, which is \( 37° + 90° = 127° \). **Explanation:** Velocity vector for point A is perpendicular to link \( O_2A \). If \( O_2A \) makes an angle θ with the x-axis, the velocity of A makes an angle of \( θ + 90° \) with the x-axis (for CCW rotation). --- ## Final Answer **The closest answer is:** **c. 127 degrees**