The speed of a train is reduced from `36 km//h` to 9` km//h` whilst it travels a distance of 50 metres, if the retardation be uniform, how much further it will travel before coming to rest ?

Answer

To solve the problem, we need to find out how much further the train will travel before coming to rest after it has already traveled 50 meters while decelerating from 36 km/h to 9 km/h. We will use the equations of motion to find the solution.

### Step-by-Step Solution:

1. **Convert the Speeds from km/h to m/s**:
- Initial speed (u) = 36 km/h = \( 36 \times \frac{5}{18} = 10 \, \text{m/s} \)
- Final speed (v) = 9 km/h = \( 9 \times \frac{5}{18} = 2.5 \, \text{m/s} \)

2. **Use the Equation of Motion to Find Retardation (a)**:
We can use the equation:
\[ v^2 = u^2 + 2as \]
where:
- \( v = 2.5 \, \text{m/s} \)
- \( u = 10 \, \text{m/s} \)
- \( s = 50 \, \text{m} \) (distance traveled while decelerating)

Rearranging the equation to find \( a \):
\[ a = \frac{v^2 - u^2}{2s} \]
Substituting the values:
\[ a = \frac{(2.5)^2 - (10)^2}{2 \times 50} \]
\[ a = \frac{6.25 - 100}{100} \]
\[ a = \frac{-93.75}{100} = -0.9375 \, \text{m/s}^2 \]

3. **Calculate the Further Distance Traveled Before Coming to Rest**:
Now we need to find out how much further the train will travel before coming to rest from the speed of 2.5 m/s. We will use the same equation of motion:
\[ v^2 = u^2 + 2as \]
Here, the final velocity \( v = 0 \) (when it comes to rest), initial velocity \( u = 2.5 \, \text{m/s} \), and we need to find \( s \) (the further distance traveled).

Rearranging the equation:
\[ 0 = (2.5)^2 + 2(-0.9375)s \]
\[ 0 = 6.25 - 1.875s \]
\[ 1.875s = 6.25 \]
\[ s = \frac{6.25}{1.875} \approx 3.33 \, \text{m} \]

### Final Answer:
The train will travel approximately **3.33 meters** further before coming to rest.

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