Math Problem Statement

A train leaves station A at 5:00 am and reaches station B at 9:00 am on the same day. Another train leaves station B at 7:00 am and reaches station A at 10:30 am on the same day. The time at which the two trains cross one another is

Anonymous Quiz

A. 8:00 am

B. 8:26 am

C. 7:26 am

D. 7:56 am

Solution

To find the time at which the two trains cross each other, we need to consider the distances, speeds, and meeting points of the trains. Here is the step-by-step approach:


Step 1: Define Variables and Calculate Speeds

  1. Train A:

    • Departure time: 5:00 am
    • Arrival time: 9:00 am
    • Total travel time: 4hours4 \, \text{hours}.
    • Distance covered = DD.
    • Speed of Train A = D4units/hour\frac{D}{4} \, \text{units/hour}.
  2. Train B:

    • Departure time: 7:00 am
    • Arrival time: 10:30 am
    • Total travel time: 3.5hours3.5 \, \text{hours}.
    • Distance covered = DD.
    • Speed of Train B = D3.5units/hour=2D7units/hour\frac{D}{3.5} \, \text{units/hour} = \frac{2D}{7} \, \text{units/hour}.

Step 2: Relative Motion and Meeting Time

  • Train A starts at 5:00 am, and Train B starts at 7:00 am. At 7:00 am, Train A has already traveled for 2hours2 \, \text{hours}: Distance traveled by Train A by 7:00 am=2×D4=D2.\text{Distance traveled by Train A by 7:00 am} = 2 \times \frac{D}{4} = \frac{D}{2}.

  • The remaining distance for Train A to cover = DD2=D2D - \frac{D}{2} = \frac{D}{2}.

  • Now, both trains are traveling towards each other. The relative speed of the two trains is: Relative speed=D4+2D7=7D+8D28=15D28units/hour.\text{Relative speed} = \frac{D}{4} + \frac{2D}{7} = \frac{7D + 8D}{28} = \frac{15D}{28} \, \text{units/hour}.

  • The time taken for the two trains to meet (from 7:00 am) is the remaining distance divided by their relative speed: Time to meet=D215D28=12×2815=1415hours.\text{Time to meet} = \frac{\frac{D}{2}}{\frac{15D}{28}} = \frac{1}{2} \times \frac{28}{15} = \frac{14}{15} \, \text{hours}.


Step 3: Convert Time to Minutes

1415hours=14×6015minutes=56minutes.\frac{14}{15} \, \text{hours} = \frac{14 \times 60}{15} \, \text{minutes} = 56 \, \text{minutes}.

  • The trains meet 56minutes56 \, \text{minutes} after 7:00 am, which is at 7:56 am.

Final Answer:

D. 7:56 am


Would you like a detailed breakdown of the speed calculations or an explanation of relative motion?

Related Questions:

  1. How is relative speed calculated for two objects moving towards each other?
  2. What is the significance of converting time into hours and minutes in these problems?
  3. How would the answer change if Train B started earlier?
  4. Can we solve this problem graphically or with equations only?
  5. How does the length of travel time affect the crossing point?

Tip:

Always calculate relative speed accurately when solving problems involving two objects moving towards each other. This simplifies the problem significantly.

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Math Problem Analysis

Mathematical Concepts

Relative Speed
Time and Distance
Linear Motion

Formulas

Speed = Distance / Time
Relative Speed = Speed1 + Speed2 (for objects moving toward each other)
Time = Distance / Speed

Theorems

Basic principles of relative motion and linear equations

Suitable Grade Level

Grades 9-10