The second's hand of a watch has length `3cm`. The speed of the end point and magnitude of change in velocity at two perpendicular positions will be

Options

A. `(pi)/(10)and(pi)/(5)cm//s`
B. `(pi)/(5)and sqrt(2)pi)/(5)cm//s`
C. `(pi)/(10)and(pi)/(5sqrt(2))cm//s`
D. `(pi)/(5sqrt(2)and(pi)/(5)cm//s`

Explanation

To solve the problem, we need to find the speed of the endpoint of the second's hand of a watch and the magnitude of change in velocity at two perpendicular positions. Let's break this down step by step.

### Step 1: Determine the Length of the Second's Hand
The length of the second's hand is given as \( r = 3 \, \text{cm} \).

### Step 2: Calculate the Angular Velocity
The second's hand completes one full revolution (which is \( 2\pi \) radians) in 60 seconds. Therefore, the angular velocity \( \omega \) can be calculated as:
\[ \omega = \frac{2\pi \, \text{radians}}{60 \, \text{seconds}} = \frac{\pi}{30} \, \text{radians/second} \]

### Step 3: Calculate the Linear Speed
The linear speed \( v \) of the endpoint of the second's hand can be calculated using the formula:
\[ v = r \cdot \omega \]
Substituting the values:
\[ v = 3 \, \text{cm} \cdot \frac{\pi}{30} \, \text{radians/second} = \frac{3\pi}{30} = \frac{\pi}{10} \, \text{cm/second} \]

### Step 4: Analyze the Change in Velocity at Two Perpendicular Positions
At two perpendicular positions, the velocities can be represented as vectors. Let’s denote the velocity at position 1 as \( \vec{v_1} \) and at position 2 as \( \vec{v_2} \). Since the magnitudes of these velocities are the same, we have:
\[ |\vec{v_1}| = |\vec{v_2}| = v = \frac{\pi}{10} \, \text{cm/second} \]

### Step 5: Calculate the Change in Velocity
The change in velocity \( \Delta \vec{v} \) can be calculated using vector subtraction:
\[ \Delta \vec{v} = \vec{v_2} - \vec{v_1} \]
Since \( \vec{v_1} \) and \( \vec{v_2} \) are perpendicular, we can use the Pythagorean theorem to find the magnitude of the change in velocity:
\[ |\Delta \vec{v}| = \sqrt{|\vec{v_1}|^2 + |\vec{v_2}|^2} = \sqrt{v^2 + v^2} = \sqrt{2v^2} = v\sqrt{2} \]
Substituting the value of \( v \):
\[ |\Delta \vec{v}| = \frac{\pi}{10} \sqrt{2} \, \text{cm/second} \]

### Step 6: Final Result
Thus, the magnitude of change in velocity at two perpendicular positions is:
\[ |\Delta \vec{v}| = \frac{\pi \sqrt{2}}{10} \, \text{cm/second} \]

### Summary of Results
- Speed of the endpoint of the second's hand: \( \frac{\pi}{10} \, \text{cm/second} \)
- Magnitude of change in velocity at two perpendicular positions: \( \frac{\pi \sqrt{2}}{10} \, \text{cm/second} \)