A particle moves along a straight line its velocity dipends on time as `v=4t-t^(2)`. Then for first `5 s`:

Options

A. Average velcotu is `25//3 m s^(-1)`
B. Average speed is `10 m s^(-1)`.
C. Average velcotu is `5//3 m s^(-1)`
D. Acceleration is `4 m s^(-2)` at `t=o`

Explanation

To solve the problem step by step, we will analyze the given velocity function \( v = 4t - t^2 \) and derive the necessary quantities such as acceleration, displacement, average velocity, and average speed.

### Step 1: Find the acceleration
The acceleration \( a \) is the derivative of velocity \( v \) with respect to time \( t \):
\[ a = \frac{dv}{dt} = \frac{d}{dt}(4t - t^2) \]
Differentiating term by term:
\[ a = 4 - 2t \]

### Step 2: Calculate acceleration at specific times
- At \( t = 0 \):
\[ a(0) = 4 - 2(0) = 4 \, \text{m/s}^2 \]
- At \( t = 4 \):
\[ a(4) = 4 - 2(4) = 4 - 8 = -4 \, \text{m/s}^2 \]

### Step 3: Find the displacement
To find the displacement \( s \), we need to integrate the velocity function:
\[ s = \int v \, dt = \int (4t - t^2) \, dt \]
Integrating term by term:
\[ s = 2t^2 - \frac{t^3}{3} + C \]
Assuming the initial displacement at \( t = 0 \) is 0, we have \( C = 0 \):
\[ s = 2t^2 - \frac{t^3}{3} \]

### Step 4: Calculate displacement from \( t = 0 \) to \( t = 5 \)
Now we substitute \( t = 5 \) into the displacement equation:
\[ s(5) = 2(5^2) - \frac{(5^3)}{3} \]
Calculating:
\[ s(5) = 2(25) - \frac{125}{3} = 50 - \frac{125}{3} \]
To combine these, convert 50 to a fraction:
\[ 50 = \frac{150}{3} \]
Thus:
\[ s(5) = \frac{150}{3} - \frac{125}{3} = \frac{25}{3} \, \text{meters} \]

### Step 5: Calculate average velocity
The average velocity \( v_{avg} \) is given by the total displacement divided by the total time:
\[ v_{avg} = \frac{s(5)}{5} = \frac{\frac{25}{3}}{5} = \frac{25}{15} = \frac{5}{3} \, \text{m/s} \]

### Summary of Results
1. Acceleration at \( t = 0 \) is \( 4 \, \text{m/s}^2 \).
2. Displacement after \( 5 \, \text{s} \) is \( \frac{25}{3} \, \text{meters} \).
3. Average velocity over the first \( 5 \, \text{s} \) is \( \frac{5}{3} \, \text{m/s} \).