The Power of the Position-Time Graph
When dealing with relative motion and time gaps, equations can sometimes become a tangled mess of variables. However, visualizing the scenario using a position-time (x−t) graph turns a complex algebraic puzzle into a simple geometry problem.
Imagine the x-axis representing time and the y-axis representing the distance along the railway track. Station A is our origin at x=0, and Station B is located at x=s=20 km. We have two trains: the passenger's train (Tp) and the first train (T1).
Analyzing the Passenger's Journey
Let's start with what we know completely: the passenger's train. It covers a distance of 20 km at an average speed of vp=60 km/h. Using the fundamental kinematic relation, we can find the time tp it takes for this journey:
tp=vps=60 km/h20 km=31 h
Converting this into minutes for easier comparison with our given data, we get tp=20 min. On our graph, the passenger's train is a straight line starting at some time tpA and ending at tpB, where the horizontal width of this line is exactly 20 min.
Connecting the Timelines
Now, let's introduce the first train, T1. We don't know its speed, so let's assume it takes a time t1 to travel between the two stations.
The announcements give us the crucial links between the two trains. At Station A, T1 passed Δt1=30 min earlier than Tp. At Station B, T1 arrived Δt2=20 min earlier than Tp.
If we look at the total time elapsed from the moment T1 leaves Station A to the moment Tp arrives at Station B, we can trace it in two distinct paths along our graph's time axis:
1. Follow T1's journey (t1) and then add the final waiting gap at Station B (Δt2).
2. Add the initial waiting gap at Station A (Δt1) and then follow Tp's journey (tp).
Equating these two paths gives us our master equation:
The Final Calculation
Rearranging the equation to solve for the unknown time t1, we get:
Substituting our known values:
t1=20 min+30 min−20 min=30 min
So, the first train took 30 min, or 0.5 h, to cover the 20 km distance. Finally, we calculate its average speed v1:
v1=t1s=0.5 h20 km=40 km/h
The General Formula:
If we substitute the symbolic expressions back into our final step, we can derive a beautiful general formula for any problem of this type:
v1=t1s=vps+Δt1−Δt2s=s+vp(Δt1−Δt2)svp
This elegant result shows how the relative time gaps directly modulate the effective speed of the leading object.