A proton in a cyclotron changes its velocity from 30 `kms^(–1)` the north of 40 `kms^(–1)` the east in 20 s. What is the magnitude of average acceleration during this time ?

Options

A. `2.5 kms^(-2)`
B. `12.5 kms^(-2)`
C. `22.5 kms^(-2)`
D. `32.5 kms^(-2)`

Answer

To find the magnitude of average acceleration of the proton, we can follow these steps:

### Step 1: Identify Initial and Final Velocities
- The initial velocity \( \vec{V_1} \) is 30 km/s to the north.
- The final velocity \( \vec{V_2} \) is 40 km/s to the east.

### Step 2: Convert Velocities to Vector Form
- We can represent the velocities as vectors:
- \( \vec{V_1} = 0 \hat{i} + 30 \hat{j} \) (where \( \hat{i} \) is the east direction and \( \hat{j} \) is the north direction)
- \( \vec{V_2} = 40 \hat{i} + 0 \hat{j} \)

### Step 3: Calculate Change in Velocity
- The change in velocity \( \Delta \vec{V} \) is given by:
\[ \Delta \vec{V} = \vec{V_2} - \vec{V_1} = (40 \hat{i} + 0 \hat{j}) - (0 \hat{i} + 30 \hat{j}) = 40 \hat{i} - 30 \hat{j} \]

### Step 4: Calculate the Magnitude of Change in Velocity
- The magnitude of the change in velocity \( |\Delta \vec{V}| \) can be calculated using the Pythagorean theorem:
\[ |\Delta \vec{V}| = \sqrt{(40)^2 + (-30)^2} = \sqrt{1600 + 900} = \sqrt{2500} = 50 \text{ km/s} \]

### Step 5: Calculate Average Acceleration
- Average acceleration \( \vec{a}_{\text{avg}} \) is defined as the change in velocity divided by the time interval:
\[ \vec{a}_{\text{avg}} = \frac{\Delta \vec{V}}{\Delta t} = \frac{50 \text{ km/s}}{20 \text{ s}} = 2.5 \text{ km/s}^2 \]

### Final Answer
- The magnitude of the average acceleration is \( 2.5 \text{ km/s}^2 \).

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