The velocity of a particle varies with time as `vecv = 3 hati + (4 - 5t)hatj ms ^(-1).` Find the average velocity of the particle for a time interval between t=0 and a time when the speed of the particle becomes minimum.

Options

A. `( 3 hati + 2 hatj ) ms ^(-1)`
B. `( 6 hati + 5 hatj ) ms ^(-1)`
C. `( 4 hati + hatj ) ms ^(-1)`
D. `( 3 hati - 5 hatj ) ms ^(-1)`

Explanation

To find the average velocity of the particle for the given time interval, we can follow these steps:

### Step 1: Identify the velocity function
The velocity of the particle is given as:
\[ \vec{v} = 3 \hat{i} + (4 - 5t) \hat{j} \, \text{m/s} \]

### Step 2: Determine when the speed is minimum
The speed of the particle is given by the magnitude of the velocity vector:
\[ |\vec{v}| = \sqrt{(3)^2 + (4 - 5t)^2} \]
To find the minimum speed, we need to minimize the expression under the square root:
\[ |\vec{v}|^2 = 9 + (4 - 5t)^2 \]
To minimize \( |\vec{v}|^2 \), we can differentiate it with respect to time \( t \) and set the derivative to zero.

### Step 3: Differentiate and find critical points
Differentiating \( |\vec{v}|^2 \):
\[ \frac{d}{dt}(9 + (4 - 5t)^2) = 2(4 - 5t)(-5) = -10(4 - 5t) \]
Setting the derivative to zero:
\[ -10(4 - 5t) = 0 \implies 4 - 5t = 0 \implies t = \frac{4}{5} \, \text{s} \]

### Step 4: Calculate the average velocity
The average velocity \( \vec{V}_{avg} \) over the time interval from \( t = 0 \) to \( t = \frac{4}{5} \) is given by:
\[ \vec{V}_{avg} = \frac{\Delta \vec{s}}{\Delta t} \]
where \( \Delta \vec{s} \) is the total displacement and \( \Delta t \) is the total time.

### Step 5: Calculate total displacement
To find the total displacement, we need to integrate the velocity vector over time:
\[ \Delta \vec{s} = \int_0^{\frac{4}{5}} \vec{v} \, dt = \int_0^{\frac{4}{5}} \left( 3 \hat{i} + (4 - 5t) \hat{j} \right) dt \]
Breaking it down:
\[ \Delta \vec{s} = \int_0^{\frac{4}{5}} 3 \hat{i} \, dt + \int_0^{\frac{4}{5}} (4 - 5t) \hat{j} \, dt \]

### Step 6: Evaluate the integrals
1. For the \( \hat{i} \) component:
\[ \int_0^{\frac{4}{5}} 3 \, dt = 3t \bigg|_0^{\frac{4}{5}} = 3 \cdot \frac{4}{5} = \frac{12}{5} \hat{i} \]

2. For the \( \hat{j} \) component:
\[ \int_0^{\frac{4}{5}} (4 - 5t) \, dt = \left( 4t - \frac{5t^2}{2} \right) \bigg|_0^{\frac{4}{5}} = \left( 4 \cdot \frac{4}{5} - \frac{5 \cdot \left(\frac{4}{5}\right)^2}{2} \right) \]
Calculating this:
\[ = \left( \frac{16}{5} - \frac{5 \cdot \frac{16}{25}}{2} \right) = \left( \frac{16}{5} - \frac{16}{10} \right) = \left( \frac{16}{5} - \frac{8}{5} \right) = \frac{8}{5} \hat{j} \]

### Step 7: Combine the components
Thus, the total displacement is:
\[ \Delta \vec{s} = \frac{12}{5} \hat{i} + \frac{8}{5} \hat{j} \]

### Step 8: Calculate the average velocity
Now, substituting into the average velocity formula:
\[ \vec{V}_{avg} = \frac{\Delta \vec{s}}{\Delta t} = \frac{\frac{12}{5} \hat{i} + \frac{8}{5} \hat{j}}{\frac{4}{5}} = \left( \frac{12}{5} \cdot \frac{5}{4} \hat{i} + \frac{8}{5} \cdot \frac{5}{4} \hat{j} \right) = \left( 3 \hat{i} + 2 \hat{j} \right) \]

### Final Answer
The average velocity of the particle is:
\[ \vec{V}_{avg} = 3 \hat{i} + 2 \hat{j} \, \text{m/s} \]