If a train takes 1.5 seconds to cross a-telegraph post and 1.75 seconds to overtake a cyclist racing along a road parallel to the track at 10 m/s, then the length of the train is

Options

A. 135 metres
B. 125 metres
C. 115 metres
D. 105 metres

Answer

To solve the problem step by step, we will follow these calculations:

### Step 1: Define Variables
Let the length of the train be \( x \) meters and the speed of the train be \( a \) meters per second.

### Step 2: Use the Time to Cross the Telegraph Post
When the train crosses a telegraph post, it travels a distance equal to its own length \( x \) in 1.5 seconds.
Using the formula for time:
\[ \text{Time} = \frac{\text{Distance}}{\text{Speed}} \]
we can write:
\[ 1.5 = \frac{x}{a} \]
Rearranging gives us:
\[ x = 1.5a \quad \text{(Equation 1)} \]

### Step 3: Use the Time to Overtake the Cyclist
The train overtakes a cyclist who is moving at 10 m/s in 1.75 seconds. The relative speed of the train with respect to the cyclist is \( a - 10 \) m/s (since they are moving in the same direction). The distance covered while overtaking the cyclist is also \( x \):
\[ 1.75 = \frac{x}{a - 10} \]
Rearranging gives us:
\[ x = 1.75(a - 10) \quad \text{(Equation 2)} \]

### Step 4: Set the Two Equations Equal
Now we have two expressions for \( x \):
\[ 1.5a = 1.75(a - 10) \]
Expanding the right side:
\[ 1.5a = 1.75a - 17.5 \]

### Step 5: Solve for \( a \)
Rearranging gives us:
\[ 1.5a - 1.75a = -17.5 \]
\[ -0.25a = -17.5 \]
Dividing both sides by -0.25:
\[ a = \frac{17.5}{0.25} = 70 \text{ m/s} \]

### Step 6: Substitute \( a \) Back to Find \( x \)
Now substitute \( a = 70 \) back into Equation 1:
\[ x = 1.5 \times 70 = 105 \text{ meters} \]

### Final Answer
The length of the train is \( 105 \) meters.

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