A particle which is initially at rest at the origin, is subjection to an acceleration with x- and y- components as shown. After time t=5 , the particle has no acceleration.

What is the magnitude of average velocity of the particle between t=0 and t=4 seconds ?

A. `(5)/(2)sqrt(13)m//s`
B. `(5)/(2)sqrt(17)m//s`
C. `30m//s`
D. None of these

Options

A. `(5)/(2)sqrt(13)m//s`
B. `(5)/(2)sqrt(17)m//s`
C. `30m//s`
D. None of these

Explanation

To solve the problem step by step, we need to find the average velocity of the particle between \( t = 0 \) and \( t = 4 \) seconds. The average velocity is defined as the total displacement divided by the total time taken.

### Step 1: Understand the problem
The particle starts from rest at the origin (0, 0). We need to calculate the displacement in both the x and y directions over the time interval from \( t = 0 \) to \( t = 4 \) seconds.

### Step 2: Determine the acceleration components
From the information provided:
- For the first 2 seconds (from \( t = 0 \) to \( t = 2 \)):
- Acceleration in the x-direction, \( a_x = 10 \, \text{m/s}^2 \)
- Acceleration in the y-direction, \( a_y = -5 \, \text{m/s}^2 \)

- For the next 2 seconds (from \( t = 2 \) to \( t = 4 \)):
- The particle continues with the velocities obtained at \( t = 2 \) seconds, but the acceleration in the x-direction is still \( a_x = 10 \, \text{m/s}^2 \) and in the y-direction is \( a_y = -5 \, \text{m/s}^2 \).

### Step 3: Calculate displacement in the x-direction
Using the equation of motion:
\[ s = ut + \frac{1}{2} a t^2 \]
For the first 2 seconds:
- Initial velocity \( u = 0 \)
- \( t = 2 \)
- \( a_x = 10 \)

Displacement in x-direction (\( x_1 \)):
\[ x_1 = 0 + \frac{1}{2} \cdot 10 \cdot (2^2) = 0 + \frac{1}{2} \cdot 10 \cdot 4 = 20 \, \text{m} \]

For the next 2 seconds (from \( t = 2 \) to \( t = 4 \)):
- Initial velocity at \( t = 2 \) is \( v_x = a_x \cdot t = 10 \cdot 2 = 20 \, \text{m/s} \)
- \( t = 2 \)
- \( a_x = 10 \)

Displacement in x-direction (\( x_2 \)):
\[ x_2 = 20 \cdot 2 + \frac{1}{2} \cdot 10 \cdot (2^2) = 40 + \frac{1}{2} \cdot 10 \cdot 4 = 40 + 20 = 60 \, \text{m} \]

Total displacement in x-direction:
\[ \Delta x = x_1 + x_2 = 20 + 60 = 80 \, \text{m} \]

### Step 4: Calculate displacement in the y-direction
For the first 2 seconds:
- Initial velocity \( u = 0 \)
- \( t = 2 \)
- \( a_y = -5 \)

Displacement in y-direction (\( y_1 \)):
\[ y_1 = 0 + \frac{1}{2} \cdot (-5) \cdot (2^2) = 0 - \frac{1}{2} \cdot 5 \cdot 4 = -10 \, \text{m} \]

For the next 2 seconds (from \( t = 2 \) to \( t = 4 \)):
- Initial velocity at \( t = 2 \) is \( v_y = -10 \, \text{m/s} \)
- \( t = 2 \)
- \( a_y = -5 \)

Displacement in y-direction (\( y_2 \)):
\[ y_2 = -10 \cdot 2 + \frac{1}{2} \cdot (-5) \cdot (2^2) = -20 - \frac{1}{2} \cdot 5 \cdot 4 = -20 - 10 = -30 \, \text{m} \]

Total displacement in y-direction:
\[ \Delta y = y_1 + y_2 = -10 - 30 = -40 \, \text{m} \]

### Step 5: Calculate the total displacement
The total displacement \( \Delta s \) is given by:
\[ \Delta s = \sqrt{(\Delta x)^2 + (\Delta y)^2} = \sqrt{(80)^2 + (-40)^2} = \sqrt{6400 + 1600} = \sqrt{8000} = 20\sqrt{20} \, \text{m} \]

### Step 6: Calculate average velocity
Average velocity \( V_{avg} \) is given by:
\[ V_{avg} = \frac{\Delta s}{\Delta t} = \frac{20\sqrt{20}}{4} = 5\sqrt{20} \, \text{m/s} \]

### Step 7: Simplify the average velocity
We can simplify \( 5\sqrt{20} \):
\[ 5\sqrt{20} = 5\sqrt{4 \cdot 5} = 5 \cdot 2\sqrt{5} = 10\sqrt{5} \, \text{m/s} \]

### Final Answer
The magnitude of the average velocity of the particle between \( t = 0 \) and \( t = 4 \) seconds is \( 10\sqrt{5} \, \text{m/s} \).