Let's analyze the problem step by step.
Problem Restatement
- Two trains, each of length 1 meter.
- They cross each other in 12 seconds.
- Speed of one train = 40 km/h.
- Find the speed of the other train.
Step 1: Convert units
Since the length is given in meters and time in seconds, convert the speed of the known train from km/h to m/s.
40 km/h=40×1000 m3600 s=400003600≈11.11 m/s40\text{ km/h}=40\times \frac{1000\text{ m}}{3600\text{ s}}=\frac{40000}{3600}\approx 11.11\text{ m/s}40 km/h=40×3600 s1000 m=360040000≈11.11 m/s
Step 2: Understand the crossing time
When two trains cross each other, the total distance covered relative to each other is the sum of their lengths.
Total length=1 m+1 m=2 m\text{Total length}=1\text{ m}+1\text{ m}=2\text{ m}Total length=1 m+1 m=2 m
They cross each other in 12 seconds, so the relative speed is:
Relative speed=Total lengthTime=2 m12 s=16 m/s≈0.1667 m/s\text{Relative speed}=\frac{\text{Total length}}{\text{Time}}=\frac{2\text{ m}}{12\text{ s}}=\frac{1}{6}\text{ m/s}\approx 0.1667\text{ m/s}Relative speed=TimeTotal length=12 s2 m=61 m/s≈0.1667 m/s
Step 3: Find the speed of the other train
Let the speed of the other train be vvv m/s. Since they are crossing each other, their relative speed is the sum of their speeds (assuming they are moving in opposite directions):
v+11.11=0.1667v+11.11=0.1667v+11.11=0.1667
This gives:
v=0.1667−11.11=−10.9433 m/sv=0.1667-11.11=-10.9433\text{ m/s}v=0.1667−11.11=−10.9433 m/s
This is negative, which is impossible for speed.
Step 4: Re-examine the problem
The relative speed cannot be less than the speed of one train alone. The total length is only 2 meters, and the crossing time is 12 seconds, which means the relative speed is very slow (0.1667 m/s). Given that one train moves at 11.11 m/s, the other train must be moving in the same direction for the relative speed to be small. Thus, if the trains move in the same direction:
Relative speed=∣v−11.11∣=0.1667 m/s\text{Relative speed}=|v-11.11|=0.1667\text{ m/s}Relative speed=∣v−11.11∣=0.1667 m/s
So,
∣v−11.11∣=0.1667|v-11.11|=0.1667∣v−11.11∣=0.1667
Two possibilities:
- v=11.11+0.1667=11.2767 m/sv=11.11+0.1667=11.2767\text{ m/s}v=11.11+0.1667=11.2767 m/s
- v=11.11−0.1667=10.9433 m/sv=11.11-0.1667=10.9433\text{ m/s}v=11.11−0.1667=10.9433 m/s
Step 5: Convert back to km/h
v1=11.2767 m/s=11.2767×36001000=40.596 km/hv_1=11.2767\text{ m/s}=11.2767\times \frac{3600}{1000}=40.596\text{ km/h}v1=11.2767 m/s=11.2767×10003600=40.596 km/h
v2=10.9433 m/s=10.9433×36001000=39.396 km/hv_2=10.9433\text{ m/s}=10.9433\times \frac{3600}{1000}=39.396\text{ km/h}v2=10.9433 m/s=10.9433×10003600=39.396 km/h
Final answer:
The speed of the other train is approximately 39.4 km/h or 40.6 km/h , depending on which train is faster. If you want me to clarify or solve a similar problem, just ask!
