To determine when the two trains are nearest to each other, we can analyze their movements and positions over time. Let’s break this down step by step.
Understanding the Scenario
We have two trains crossing at a junction:
- Train A travels at 30 km/h and crosses the junction at 7 PM.
- Train B travels at 40 km/h and crosses the junction at midnight.
Setting Up the Problem
First, we need to establish a coordinate system. Let’s assume the junction is at the origin (0, 0) of a Cartesian plane:
- Train A moves along the x-axis.
- Train B moves along the y-axis.
Position of Each Train Over Time
Next, we can express the positions of the trains as functions of time:
Position of Train A
Train A starts at the junction at 7 PM. If we let \( t \) represent the time in hours after 7 PM, the position of Train A at time \( t \) is:
X_A(t) = 30t (in km)
Position of Train B
Train B starts at the junction at midnight, which is 5 hours after Train A. Therefore, if \( t \) is the time in hours after 7 PM, Train B's position is given by:
X_B(t) = 40(t - 5) (for \( t \geq 5 \), in km)
Finding the Distance Between the Trains
The distance \( D \) between the two trains at any time \( t \) can be calculated using the distance formula:
D(t) = √[(X_A(t))^2 + (X_B(t))^2]
Substituting the positions of the trains:
D(t) = √[(30t)^2 + (40(t - 5))^2]
Minimizing the Distance
To find when the trains are nearest, we need to minimize \( D(t) \). Instead of minimizing \( D(t) \) directly, we can minimize \( D^2(t) \) to simplify calculations:
D^2(t) = (30t)^2 + (40(t - 5))^2
Expanding this gives:
D^2(t) = 900t^2 + 1600(t^2 - 10t + 25)
D^2(t) = 900t^2 + 1600t^2 - 16000t + 40000
D^2(t) = 2500t^2 - 16000t + 40000
Finding the Minimum Value
To find the minimum, we can use the vertex formula for a quadratic equation \( at^2 + bt + c \), where the vertex occurs at \( t = -\frac{b}{2a} \):
Here, \( a = 2500 \) and \( b = -16000 \):
t = -\frac{-16000}{2 \times 2500} = \frac{16000}{5000} = 3.2
Interpreting the Result
This means the trains are nearest to each other 3.2 hours after 7 PM:
7 PM + 3.2 hours = 10:12 PM
Final Thoughts
Thus, the two trains will be closest to each other at approximately 10:12 PM. This approach illustrates how we can use algebra and geometry to solve real-world problems involving motion and distance.
