A train accelerating uniormly from rest attains a maximum speed of `40ms^(-1)` in `20s`. It travels at this speed for `20s` and is brought to rest with uniform retardation i further `40s`. What is the average velocity during this period?

Options

A. `80//3ms^(-1)`
B. `40 ms^(-1)`
C. `25 ms^(-1)`
D. `30 ms^(-1)`

Explanation

To solve the problem step by step, we will calculate the average velocity of the train during its entire journey.

### Step 1: Identify the phases of motion
The train goes through three distinct phases:
1. **Acceleration from rest to maximum speed** (Phase 1)
2. **Constant speed** (Phase 2)
3. **Deceleration to rest** (Phase 3)

### Step 2: Calculate the total time
The total time for the journey is the sum of the time for each phase:
- Time for Phase 1 (t1) = 20 seconds
- Time for Phase 2 (t2) = 20 seconds
- Time for Phase 3 (t3) = 40 seconds

Total time (T) = t1 + t2 + t3 = 20 + 20 + 40 = **80 seconds**

### Step 3: Calculate the displacement for each phase

#### Phase 1: Acceleration
- Initial velocity (u) = 0 m/s
- Final velocity (v) = 40 m/s
- Time (t1) = 20 s

Using the formula for acceleration:
\[ v = u + at \]
\[ 40 = 0 + a \cdot 20 \]
\[ a = \frac{40}{20} = 2 \, \text{m/s}^2 \]

Now, calculate the displacement (S1) during this phase using:
\[ S = ut + \frac{1}{2} a t^2 \]
\[ S1 = 0 \cdot 20 + \frac{1}{2} \cdot 2 \cdot (20)^2 \]
\[ S1 = 0 + \frac{1}{2} \cdot 2 \cdot 400 \]
\[ S1 = 400 \, \text{meters} \]

#### Phase 2: Constant Speed
- Speed = 40 m/s
- Time (t2) = 20 s

Displacement (S2) during this phase:
\[ S2 = \text{speed} \cdot \text{time} \]
\[ S2 = 40 \cdot 20 = 800 \, \text{meters} \]

#### Phase 3: Deceleration
- Initial velocity (u) = 40 m/s
- Final velocity (v) = 0 m/s
- Time (t3) = 40 s

Using the formula for deceleration:
\[ v = u + at \]
\[ 0 = 40 + a \cdot 40 \]
\[ a = \frac{-40}{40} = -1 \, \text{m/s}^2 \]

Now, calculate the displacement (S3) during this phase:
\[ S3 = ut + \frac{1}{2} a t^2 \]
\[ S3 = 40 \cdot 40 + \frac{1}{2} \cdot (-1) \cdot (40)^2 \]
\[ S3 = 1600 - \frac{1}{2} \cdot 1600 \]
\[ S3 = 1600 - 800 = 800 \, \text{meters} \]

### Step 4: Calculate total displacement
Total displacement (S) = S1 + S2 + S3
\[ S = 400 + 800 + 800 = 2000 \, \text{meters} \]

### Step 5: Calculate average velocity
Average velocity (V_avg) is given by:
\[ V_{avg} = \frac{\text{Total Displacement}}{\text{Total Time}} \]
\[ V_{avg} = \frac{2000}{80} = 25 \, \text{m/s} \]

### Final Answer
The average velocity of the train during this period is **25 m/s**.

---