A

Step 1: Determine the initial and final velocities of the tip of the second's hand.
The second's hand rotates in a circle with a radius of 1 cm. The angular velocity of the second's hand, which completes one full revolution in 60 seconds, is given by:
$$\omega = \frac{2\pi \text{ rad}}{60 \text{ s}} = \frac{\pi}{30} \text{ rad/s}$$
Step 2: Calculate the velocity (v) of the tip at any instant using the formula:
$$v = r \cdot \omega$$
where r is the radius (1 cm or 0.01 m). Thus,
$$v = 0.01 \text{ m} \cdot \frac{\pi}{30} \text{ rad/s} = \frac{\pi}{3000} \text{ m/s}$$
Step 3: After 15 seconds, the hand completes a half revolution (180 degrees or π radians), resulting in a change in direction of the velocity vector, while the speed (magnitude of velocity) remains the same.
Initially, the velocity vector can be taken as pointing to the right (0 degrees) and after 15 seconds, it points directly downward (180 degrees).
Step 4: The change in velocity can be calculated using the formula for change in vector quantities:
Let $$\vec{v_1} = \frac{\pi}{3000} \hat{i}$$
and $$\vec{v_2} = -\frac{\pi}{3000} \hat{j}$$
The change in velocity $$\Delta \vec{v} = \vec{v_2} - \vec{v_1} = -\frac{\pi}{3000} \hat{j} - \frac{\pi}{3000} \hat{i} = -\frac{\pi}{3000} (\hat{i} + \hat{j})$$
The magnitude of change in velocity is:
$$|\Delta \vec{v}| = \sqrt{\left(-\frac{\pi}{3000}\right)^2 + \left(-\frac{\pi}{3000}\right)^2} = \frac{\pi}{3000}\sqrt{2}$$
Step 5: Now, calculate the average acceleration during this interval:
Average acceleration ($a_{avg}$) is given by:
$$a_{avg} = \frac{\Delta v}{\Delta t}$$
where $$\Delta t = 15 s$$. Thus,
$$a_{avg} = \frac{|\Delta \vec{v}|}{15}\text{ s} = \frac{\frac{\pi}{3000}\sqrt{2}}{15} = \frac{\pi \sqrt{2}}{45000} \text{ m/s}^2$$
Final Result:
Magnitude of change in velocity: $$\frac{\pi}{3000}\sqrt{2} \text{ m/s}$$
Magnitude of average acceleration: $$\frac{\pi \sqrt{2}}{45000} \text{ m/s}^2$$.