Calculating Train Speed from Crossing Times

This problem involves understanding how the length of a train and the length of an object it crosses affect the time taken at a constant speed. When a train crosses a point object like a pole, it covers a distance equal to its own length. When it crosses an extended object like a bridge, it covers a distance equal to its own length plus the length of the bridge.

Understanding the Concepts

• When a train crosses a pole, the distance covered by the train is equal to the length of the train itself. The time taken is related by the formula: Distance = Speed × Time.
• When a train crosses a bridge, the distance covered by the train is equal to the length of the train plus the length of the bridge. The time taken is also related by the formula: Distance = Speed × Time.

Setting Up the Equations

Let:

• $L_t$ be the length of the train (in meters)
• $v$ be the speed of the train (in meters per second, m/s)
• $t_p$ be the time taken to cross the pole (in seconds)
• $L_b$ be the length of the bridge (in meters)
• $t_b$ be the time taken to cross the bridge (in seconds)

From the problem statement, we have:

• $t_p = 12$ seconds
• $L_b = 170$ meters
• $t_b = 36$ seconds

Using the distance, speed, and time relationship:

For crossing the pole:

Distance = Train Length

$L_t = v \times t_p$

$L_t = v \times 12 \quad (Equation\ 1)$

For crossing the bridge:

Distance = Train Length + Bridge Length

$L_t + L_b = v \times t_b$

$L_t + 170 = v \times 36 \quad (Equation\ 2)$

Solving for the Train Speed

We have two equations and two unknowns ($L_t$ and $v$). We can substitute Equation 1 into Equation 2 to eliminate $L_t$ and solve for $v$.

Substitute $L_t = 12v$ into Equation 2:

$(12v) + 170 = 36v$

Now, we solve for $v$:

Subtract $12v$ from both sides:

$170 = 36v - 12v$

$170 = 24v$

Divide by 24 to find $v$:

$v = \frac{170}{24}$ m/s

We can simplify the fraction:

$v = \frac{85}{12}$ m/s

Converting Speed to km/h

The options are given in kilometers per hour (km/h). To convert speed from meters per second (m/s) to kilometers per hour (km/h), we multiply by $\frac{18}{5}$.

$v_{km/h} = v_{m/s} \times \frac{18}{5}$

$v_{km/h} = \frac{85}{12} \times \frac{18}{5}$

Let's simplify the calculation:

$v_{km/h} = \left(\frac{85}{5}\right) \times \left(\frac{18}{12}\right)$

$v_{km/h} = 17 \times \frac{3}{2}$

$v_{km/h} = \frac{51}{2}$

$v_{km/h} = 25.5$ km/h

Thus, the speed of the train is 25.5 km/h.

Checking the Options

Comparing our calculated speed with the given options:

• 30.75 km/h
• 25.5 km/h
• 32.45 km/h
• 10.8 km/h

Our calculated speed of 25.5 km/h matches one of the options.

Final Answer

The speed of the train is 25.5 km/h.

Revision Table: Train Speed Calculation

Additional Information: Speed, Distance, and Time

The relationship between speed, distance, and time is fundamental in physics and mathematics problems involving motion. The basic formula is:

$\text{Speed} = \frac{\text{Distance}}{\text{Time}}$

This formula can be rearranged to find Distance or Time if the other two quantities are known:

• $\text{Distance} = \text{Speed} \times \text{Time}$
• $\text{Time} = \frac{\text{Distance}}{\text{Speed}}$

It is crucial to ensure that the units are consistent when using these formulas. If distance is in meters and time is in seconds, the speed will be in meters per second (m/s). If distance is in kilometers and time is in hours, the speed will be in kilometers per hour (km/h).

Conversion factors are often needed when units are mixed. The most common conversion for speed is between m/s and km/h:

• To convert m/s to km/h, multiply by $\frac{18}{5}$. (Since 1 km = 1000 m and 1 hour = 3600 seconds, $1 \text{ m/s} = \frac{1 \text{ m}}{1 \text{ s}} = \frac{1/1000 \text{ km}}{1/3600 \text{ h}} = \frac{1}{1000} \times 3600 \text{ km/h} = \frac{3600}{1000} \text{ km/h} = \frac{18}{5} \text{ km/h}$).
• To convert km/h to m/s, multiply by $\frac{5}{18}$.