A

Step 1: Calculate the angular displacement in radians. Since the second's hand completes a full rotation (360 degrees) in 60 seconds, the angular velocity \( \omega \) can be calculated as:
\( \omega = \frac{2\pi \text{ radians}}{60 ext{ seconds}} = \frac{\pi}{30} \text{ radians/second} \).

Step 2: Find the angular displacement in 15 seconds. This can be calculated as:
\( \theta = \omega \cdot t = \frac{\pi}{30} imes 15 = \frac{\pi}{2} \text{ radians} \).

Step 3: Calculate the initial and final velocities of the tip of the second's hand. The radius \( r = 1 \text{ cm} \). Using the formula for linear velocity \( v = r\omega \), we find:
Initial velocity \( v_i = 1 \cdot \frac{\pi}{30} ext{ cm/s} \) and after the change, the final velocity is at an angle of \( \frac{\pi}{2} \), so:
Final velocity \( v_f = 1 \cdot \frac{\pi}{30} ext{ cm/s} \) but perpendicular to the initial velocity.

Step 4: The magnitude of change in velocity can be derived using the vector form. The change in velocity is not simply the difference in speed but also accounts for the change in direction.
Therefore, using vector addition: \( |\Delta v| = v_f - v_i = \sqrt{(v_f^2) + (v_i^2)} = \sqrt{(\frac{\pi}{30})^2 + (\frac{\pi}{30})^2} = \sqrt{2(\frac{\pi^2}{900})} = \frac{\pi}{30}\sqrt{2} \text{ cm/s} , \approx 0.147 ext{ cm/s} \).

Therefore, the magnitude of change in velocity of the tip of the second's hand in 15 seconds is approximately 0.146 cm/s.