The speed of a train is reduced from 60 `(km)/h` to 15 `(km)/h` whilst it travels a distance of 450 meters, if the retardation be uniform, how much distance it will travel before coming to rest.

Answer

To solve the problem, we will follow these steps:

### Step 1: Convert the speeds from km/h to m/s
The initial speed \( u \) is given as 60 km/h. To convert this to meters per second (m/s), we use the conversion factor \( \frac{5}{18} \).

\[ u = 60 \, \text{km/h} \times \frac{5}{18} = \frac{300}{18} = 16.67 \, \text{m/s} \]

The final speed \( v \) is given as 15 km/h. Similarly, we convert this to m/s:

\[ v = 15 \, \text{km/h} \times \frac{5}{18} = \frac{75}{18} = 4.17 \, \text{m/s} \]

### Step 2: Use the equation of motion to find retardation
We know the initial velocity \( u \), final velocity \( v \), and the distance \( s \) traveled during the deceleration. We can use the equation:

\[ v^2 = u^2 + 2as \]

Rearranging for \( a \) (acceleration or retardation in this case):

\[ a = \frac{v^2 - u^2}{2s} \]

Substituting the values we have:

\[ a = \frac{(4.17)^2 - (16.67)^2}{2 \times 450} \]

Calculating \( (4.17)^2 \) and \( (16.67)^2 \):

\[ (4.17)^2 = 17.39 \quad \text{and} \quad (16.67)^2 = 278.89 \]

Now substituting these values into the equation:

\[ a = \frac{17.39 - 278.89}{900} = \frac{-261.5}{900} \approx -0.2917 \, \text{m/s}^2 \]

### Step 3: Calculate the distance traveled before coming to rest
Now we need to find the distance \( s_1 \) the train will travel before coming to rest from the speed of 15 km/h (4.17 m/s) to 0 m/s. We can use the same equation of motion:

\[ v^2 = u^2 + 2as_1 \]

Here, \( v = 0 \), \( u = 4.17 \, \text{m/s} \), and \( a = -0.2917 \, \text{m/s}^2 \).

Rearranging gives:

\[ 0 = (4.17)^2 + 2(-0.2917)s_1 \]

Solving for \( s_1 \):

\[ s_1 = \frac{(4.17)^2}{2 \times 0.2917} \]

Calculating:

\[ s_1 = \frac{17.39}{0.5834} \approx 29.84 \, \text{meters} \]

### Final Answer
The distance the train will travel before coming to rest is approximately **29.84 meters**.

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