A particle moves in the direction of east for 2s with velocity of `15 ms^(-1).` Then it moves towards north for 8s with a a velocity of `5 ms^(-1).` The average velocity of the particle is

Options

A. `1ms ^(-1)`
B. `5ms ^(-1)`
C. `7ms^(-1)`
D. `10ms^(-1)`

Answer

To find the average velocity of the particle, we will follow these steps:

### Step 1: Calculate the distance traveled in the east direction
The particle moves east with a velocity of \(15 \, \text{m/s}\) for \(2 \, \text{s}\).

\[ \text{Distance in east direction} (x_1) = \text{velocity} \times \text{time} = 15 \, \text{m/s} \times 2 \, \text{s} = 30 \, \text{m} \]

### Step 2: Calculate the distance traveled in the north direction
The particle then moves north with a velocity of \(5 \, \text{m/s}\) for \(8 \, \text{s}\).

\[ \text{Distance in north direction} (x_2) = \text{velocity} \times \text{time} = 5 \, \text{m/s} \times 8 \, \text{s} = 40 \, \text{m} \]

### Step 3: Calculate the total displacement using Pythagorean theorem
The total displacement (\(\Delta x\)) can be calculated using the Pythagorean theorem since the motion in east and north forms a right triangle.

\[ \Delta x = \sqrt{x_1^2 + x_2^2} = \sqrt{(30 \, \text{m})^2 + (40 \, \text{m})^2} = \sqrt{900 + 1600} = \sqrt{2500} = 50 \, \text{m} \]

### Step 4: Calculate the total time taken
The total time (\(\Delta t\)) taken for the entire motion is the sum of the time taken in each direction.

\[ \Delta t = 2 \, \text{s} + 8 \, \text{s} = 10 \, \text{s} \]

### Step 5: Calculate the average velocity
The average velocity (\(V_{avg}\)) is given by the total displacement divided by the total time taken.

\[ V_{avg} = \frac{\Delta x}{\Delta t} = \frac{50 \, \text{m}}{10 \, \text{s}} = 5 \, \text{m/s} \]

### Final Answer
The average velocity of the particle is \(5 \, \text{m/s}\).

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