An `alpha` - particle in a cyclotron changes its velocity from `30` km/s south to `40` km/s west in `10` second what is the magnitude of average acceleration during this time :

Options

A. `5"Km/s"^(2)`
B. `7"Km/s"^(2)`
C. `9"Km/s"^(2)`
D. `11"Km/s"^(2)`

Explanation

To solve the problem of finding the magnitude of average acceleration of an alpha particle changing its velocity, we can follow these steps:

### Step 1: Identify Initial and Final Velocities
- The initial velocity \( v_1 \) is given as \( 30 \) km/s south. In vector form, this can be represented as:
\[ v_1 = -30 \hat{j} \text{ km/s} \]
(where \(\hat{j}\) represents the positive y-direction and negative indicates south).

- The final velocity \( v_2 \) is given as \( 40 \) km/s west. In vector form, this can be represented as:
\[ v_2 = -40 \hat{i} \text{ km/s} \]
(where \(\hat{i}\) represents the positive x-direction and negative indicates west).

### Step 2: Calculate the Change in Velocity
- The change in velocity \( \Delta v \) can be calculated as:
\[ \Delta v = v_2 - v_1 \]
Substituting the values:
\[ \Delta v = (-40 \hat{i}) - (-30 \hat{j}) = -40 \hat{i} + 30 \hat{j} \]

### Step 3: Calculate Average Acceleration
- Average acceleration \( a_{avg} \) is defined as the change in velocity divided by the time interval \( t \):
\[ a_{avg} = \frac{\Delta v}{t} \]
Given that \( t = 10 \) seconds, we can substitute:
\[ a_{avg} = \frac{-40 \hat{i} + 30 \hat{j}}{10} = -4 \hat{i} + 3 \hat{j} \text{ km/s}^2 \]

### Step 4: Calculate the Magnitude of Average Acceleration
- The magnitude of the average acceleration vector can be calculated using the Pythagorean theorem:
\[ |a_{avg}| = \sqrt{(-4)^2 + (3)^2} \]
Calculating this gives:
\[ |a_{avg}| = \sqrt{16 + 9} = \sqrt{25} = 5 \text{ km/s}^2 \]

### Conclusion
- The magnitude of the average acceleration of the alpha particle is:
\[ 5 \text{ km/s}^2 \]