Math Problem Statement

. Two trains are traveling in the opposite direction at uniform speeds of 60 and 50 km per hour respectively. They take 5 seconds to cross each other. If the two trains had travelled in the same direction, then a passenger sitting in the faster-moving train would have overtaken the other train in 18 seconds. What are the lengths of trains (in metres)?

(a) 112, 78

(b) 97.78, 55

(c) 102.78, 50

(d) 102.78, 55

Solution

To solve this problem, we'll use the information given and apply the concepts of relative speed and the formula for speed, distance, and time.

Step 1: Analyze the scenario where the trains travel in opposite directions.

When two trains are traveling in opposite directions, their relative speed is the sum of their individual speeds.

Relative speed=60 km/h+50 km/h=110 km/h\text{Relative speed} = 60 \text{ km/h} + 50 \text{ km/h} = 110 \text{ km/h}

Convert this speed to meters per second (m/s) since the time is given in seconds:

110 km/h=110×10003600 m/s=1100003600 m/s30.56 m/s110 \text{ km/h} = \frac{110 \times 1000}{3600} \text{ m/s} = \frac{110000}{3600} \text{ m/s} \approx 30.56 \text{ m/s}

Since they take 5 seconds to cross each other, the combined length of the two trains L1+L2L_1 + L_2 is:

L1+L2=Relative speed×time=30.56 m/s×5 s152.78 metersL_1 + L_2 = \text{Relative speed} \times \text{time} = 30.56 \text{ m/s} \times 5 \text{ s} \approx 152.78 \text{ meters}

Step 2: Analyze the scenario where the trains travel in the same direction.

When the trains are traveling in the same direction, their relative speed is the difference between their speeds:

Relative speed=60 km/h50 km/h=10 km/h\text{Relative speed} = 60 \text{ km/h} - 50 \text{ km/h} = 10 \text{ km/h}

Convert this speed to meters per second:

10 km/h=10×10003600 m/s2.78 m/s10 \text{ km/h} = \frac{10 \times 1000}{3600} \text{ m/s} \approx 2.78 \text{ m/s}

In this case, the length of the slower train L2L_2 will be overtaken by the faster train in 18 seconds. Thus:

L2=Relative speed×time=2.78 m/s×18 s50 metersL_2 = \text{Relative speed} \times \text{time} = 2.78 \text{ m/s} \times 18 \text{ s} \approx 50 \text{ meters}

Step 3: Determine the length of the faster train L1L_1.

Since L1+L2=152.78L_1 + L_2 = 152.78 meters and L250L_2 \approx 50 meters:

L1=152.7850=102.78 metersL_1 = 152.78 - 50 = 102.78 \text{ meters}

Conclusion:

The lengths of the two trains are approximately:

L1=102.78 meters,L2=50 metersL_1 = 102.78 \text{ meters}, \quad L_2 = 50 \text{ meters}

Thus, the correct answer is:

(c) 102.78, 50

Would you like more details or have any questions? Here are some related questions to deepen your understanding:

  1. How do you convert km/h to m/s?
  2. What is relative speed and how is it used in train problems?
  3. Why does the relative speed change depending on the direction of travel?
  4. Can the same methods be applied to problems involving boats or planes?
  5. How would the problem change if the time to overtake in the same direction was different?

Tip: Always ensure units are consistent when solving problems, particularly when time is given in seconds and speed in km/h.

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Math Problem Analysis

Mathematical Concepts

Relative Speed
Uniform Motion
Unit Conversion

Formulas

Relative speed = Speed of train 1 + Speed of train 2 (opposite directions)
Distance = Speed × Time
Relative speed = Speed of train 1 - Speed of train 2 (same direction)
Speed conversion: 1 km/h = (1000/3600) m/s

Theorems

Relative speed theorem in uniform motion

Suitable Grade Level

Grades 10-12