Given the lengths of the two trains: $L_1 = 200$ m and $L_2 = 400$ m. Let their speeds be $S_1$ and $S_2$ (in m/s), with $S_1$ being the higher speed.
The total distance $D$ the trains cover relative to each other is the sum of their lengths:
$D = L_1 + L_2 = 200 \text{ m} + 400 \text{ m} = 600 \text{ m}$1. Same Direction:
When traveling in the same direction, the relative speed is $S_{rel\_same} = S_1 - S_2$. The time taken to overtake is $t_{same} = 30$ seconds.
Applying the distance formula ($D = \text{speed} \times \text{time}$):
$600 \text{ m} = (S_1 - S_2) \times 30 \text{ s}$Calculating the relative speed:
$S_1 - S_2 = \frac{600}{30} \text{ m/s} = 20 \text{ m/s}$2. Opposite Directions:
When traveling in opposite directions, the relative speed is $S_{rel\_opp} = S_1 + S_2$. The time taken to cross is $t_{opp} = 6$ seconds.
Applying the distance formula:
$600 \text{ m} = (S_1 + S_2) \times 6 \text{ s}$Calculating the relative speed:
$S_1 + S_2 = \frac{600}{6} \text{ m/s} = 100 \text{ m/s}$We have the following system of equations:
Adding Equation 1 and Equation 2:
$(S_1 - S_2) + (S_1 + S_2) = 20 + 100$ $2S_1 = 120 \text{ m/s}$ $S_1 = 60 \text{ m/s}$Substituting $S_1 = 60$ m/s into Equation 2:
$60 \text{ m/s} + S_2 = 100 \text{ m/s}$ $S_2 = 40 \text{ m/s}$To convert speeds from meters per second (m/s) to kilometers per hour (km/h), multiply by the conversion factor $\frac{18}{5}$.
Higher speed ($S_1$):
$S_1 = 60 \times \frac{18}{5} \text{ km/h} = 12 \times 18 \text{ km/h} = 216 \text{ km/h}$Lower speed ($S_2$):
$S_2 = 40 \times \frac{18}{5} \text{ km/h} = 8 \times 18 \text{ km/h} = 144 \text{ km/h}$The speeds of the two trains are 216 km/h and 144 km/h.
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