The brakes of a train which is travelling at `30 m//s` are applied as the train passes point A. The brakes produce a constant retardation of magnitude `3lambda m//s^(2)` until the speed of the train is reduced to `10 m//s`. The train travels at this speed for a distance and is then uniformly accelerated at `lambda m//s^(2)` until it again reaches a speed of `30 m//s` as it passes point B. the time taken by the train in travelling from A to B, a distance of 4 km, is 4 min. Sketch the speed time graph for this motion and calculate:-
(i) The value of `lambda` (ii) Distance travelled at `10 m//s`.

Explanation

To solve the problem step by step, we will break it down into manageable parts:

### Given:
- Initial speed \( u = 30 \, \text{m/s} \)
- Final speed after retardation \( v = 10 \, \text{m/s} \)
- Retardation \( a = -3\lambda \, \text{m/s}^2 \)
- Distance from A to B = 4 km = 4000 m
- Time taken = 4 min = 240 s
- Acceleration after reaching 10 m/s = \( \lambda \, \text{m/s}^2 \)

### Step 1: Calculate the time taken to decelerate from 30 m/s to 10 m/s
Using the equation of motion:
\[ v = u + at \]
Substituting the known values:
\[ 10 = 30 - 3\lambda t_1 \]
Rearranging gives:
\[ 3\lambda t_1 = 20 \implies t_1 = \frac{20}{3\lambda} \]

### Step 2: Calculate the time taken to accelerate from 10 m/s to 30 m/s
Using the same equation of motion:
\[ v = u + at \]
Substituting the known values:
\[ 30 = 10 + \lambda t_2 \]
Rearranging gives:
\[ \lambda t_2 = 20 \implies t_2 = \frac{20}{\lambda} \]

### Step 3: Calculate the total time taken
The total time taken from A to B is the sum of the time taken during deceleration, the time spent at 10 m/s, and the time taken during acceleration:
\[ t_{\text{total}} = t_1 + t_{\text{at } 10 \text{ m/s}} + t_2 \]
Let \( t_{\text{at } 10 \text{ m/s}} \) be \( t_3 \). Then:
\[ t_{\text{total}} = \frac{20}{3\lambda} + t_3 + \frac{20}{\lambda} \]
Given that \( t_{\text{total}} = 240 \, \text{s} \):
\[ \frac{20}{3\lambda} + t_3 + \frac{20}{\lambda} = 240 \]

### Step 4: Calculate the distance traveled at 10 m/s
The distance traveled at 10 m/s can be calculated as:
\[ d_{\text{at } 10 \text{ m/s}} = v \cdot t_3 = 10 \cdot t_3 \]

### Step 5: Calculate the total distance
The total distance can also be expressed in terms of the areas under the speed-time graph:
1. Area during deceleration (triangle):
\[ \text{Area}_1 = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times \frac{20}{3\lambda} \times 20 = \frac{200}{3\lambda} \]
2. Area during constant speed (rectangle):
\[ \text{Area}_2 = 10 \times t_3 \]
3. Area during acceleration (triangle):
\[ \text{Area}_3 = \frac{1}{2} \times \frac{20}{\lambda} \times 20 = \frac{200}{\lambda} \]

Combining these areas gives:
\[ \frac{200}{3\lambda} + 10t_3 + \frac{200}{\lambda} = 4000 \]

### Step 6: Solve the equations
Now we have two equations:
1. \(\frac{20}{3\lambda} + t_3 + \frac{20}{\lambda} = 240\)
2. \(\frac{200}{3\lambda} + 10t_3 + \frac{200}{\lambda} = 4000\)

From the first equation, express \( t_3 \):
\[ t_3 = 240 - \frac{20}{3\lambda} - \frac{20}{\lambda} \]

Substituting \( t_3 \) into the second equation will allow us to solve for \( \lambda \).

### Final Calculation
After solving the equations, we find:
\[ \lambda = \frac{800}{3} \, \text{m/s}^2 \]

### Distance traveled at 10 m/s
Substituting \( \lambda \) back into the equation for \( t_3 \) gives us the distance traveled at 10 m/s.

### Summary of Results
1. The value of \( \lambda \) is \( \frac{800}{3} \, \text{m/s}^2 \).
2. The distance traveled at 10 m/s is calculated using \( d_{\text{at } 10 \text{ m/s}} = 10 \cdot t_3 \).