A particle moves along a straight line and its velocity depends on time as `v = 3t - t^2.` Here, v is in `m//s` and t in second. Find
(a) average velocity and
(b) average speed for first five seconds.

Explanation

To solve the problem, we need to find the average velocity and average speed of a particle whose velocity is given by the equation \( v = 3t - t^2 \) over the time interval from \( t = 0 \) to \( t = 5 \) seconds.

### Step-by-Step Solution

**Step 1: Find the displacement over the time interval.**

The displacement \( s \) can be found by integrating the velocity function with respect to time:

\[ s = \int_{0}^{5} v \, dt = \int_{0}^{5} (3t - t^2) \, dt \]

**Step 2: Perform the integration.**

Calculating the integral:

\[ s = \int (3t - t^2) \, dt = \frac{3t^2}{2} - \frac{t^3}{3} + C \]

Now, we evaluate this from \( t = 0 \) to \( t = 5 \):

\[ s = \left[ \frac{3(5)^2}{2} - \frac{(5)^3}{3} \right] - \left[ \frac{3(0)^2}{2} - \frac{(0)^3}{3} \right] \]

Calculating the values:

\[ s = \left[ \frac{3 \cdot 25}{2} - \frac{125}{3} \right] \]

Calculating \( \frac{3 \cdot 25}{2} = 37.5 \) and \( \frac{125}{3} \approx 41.67 \):

\[ s = 37.5 - 41.67 = -4.17 \, \text{meters} \]

**Step 3: Calculate the average velocity.**

The average velocity \( v_{avg} \) is given by:

\[ v_{avg} = \frac{\text{Total Displacement}}{\text{Total Time}} = \frac{s}{t} \]

Substituting the values:

\[ v_{avg} = \frac{-4.17}{5} = -0.834 \, \text{m/s} \]

**Step 4: Find the total distance traveled.**

To find the average speed, we need to calculate the total distance traveled. First, we need to find when the particle changes direction by setting the velocity to zero:

\[ 3t - t^2 = 0 \implies t(3 - t) = 0 \]

This gives us \( t = 0 \) and \( t = 3 \) seconds. The particle changes direction at \( t = 3 \) seconds.

**Step 5: Calculate the distance from \( t = 0 \) to \( t = 3 \).**

\[ d_1 = \int_{0}^{3} (3t - t^2) \, dt \]

Calculating this integral:

\[ d_1 = \left[ \frac{3(3)^2}{2} - \frac{(3)^3}{3} \right] = \left[ \frac{27}{2} - 9 \right] = \left[ 13.5 - 9 \right] = 4.5 \, \text{meters} \]

**Step 6: Calculate the distance from \( t = 3 \) to \( t = 5 \).**

For \( t = 3 \) to \( t = 5 \), the velocity is negative, so we need to calculate the distance using the absolute value of the velocity:

\[ d_2 = \int_{3}^{5} |v| \, dt = \int_{3}^{5} (t^2 - 3t) \, dt \]

Calculating this integral:

\[ d_2 = \left[ \frac{t^3}{3} - \frac{3t^2}{2} \right]_{3}^{5} \]

Calculating the values:

\[ d_2 = \left[ \frac{(5)^3}{3} - \frac{3(5)^2}{2} \right] - \left[ \frac{(3)^3}{3} - \frac{3(3)^2}{2} \right] \]

Calculating:

\[ d_2 = \left[ \frac{125}{3} - \frac{75}{2} \right] - \left[ 9 - \frac{27}{2} \right] \]

Calculating \( \frac{125}{3} \approx 41.67 \) and \( \frac{75}{2} = 37.5 \):

\[ d_2 = 41.67 - 37.5 - (9 - 13.5) = 4.17 + 4.5 = 8.67 \, \text{meters} \]

**Step 7: Calculate the total distance and average speed.**

Total distance \( D \):

\[ D = d_1 + d_2 = 4.5 + 8.67 = 13.17 \, \text{meters} \]

Average speed \( v_{avg, speed} \):

\[ v_{avg, speed} = \frac{D}{t} = \frac{13.17}{5} = 2.634 \, \text{m/s} \]

### Final Answers
- **Average Velocity:** \( -0.834 \, \text{m/s} \)
- **Average Speed:** \( 2.634 \, \text{m/s} \)