A train leaves Station A at 60 mph. Two hours later, a faster train leaves Station A on the same track at 90 mph. How many hours after the first train departs does the second train catch up? - Treasure Valley Movers

How Long Does It Take for a Faster Train to Catch a Slower One? A Step-by-Step Breakdown

When two trains travel the same route on identical tracks, the pursuit becomes a classic example of relative motion. In this scenario, a train departs Station A at 60 miles per hour (mph), and two hours later, a faster train leaves the same station traveling at 90 mph. The question is: how long after the first train’s departure does the second train catch up?

Let’s break it down using basic physics and distance-time relationships.

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The Head Start Problem

The key issue is the two-hour head start the slower train has. While the first train moves steadily at 60 mph, by the time the second train starts two hours later, the first train has already traveled a significant distance.

In those two hours:

• Distance covered by Train A = Speed × Time = 60 mph × 2 hours = 120 miles

So, when Train B departs, Train A is already 120 miles ahead.

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Relative Speed: The Race Begins

Now, both trains travel on the same track. The second train moves faster—90 mph—compared to Train A’s 60 mph. The relative speed, which measures how fast the second train reduces the gap, is:

\[
90 \, \ ext{mph} - 60 \, \ ext{mph} = 30 \, \ ext{mph}
\]

This means Train B gains on Train A by 30 miles every hour after the second train starts.

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How Long to Close the 120-Mile Gap

To catch up, Train B must cover the initial 120-mile deficit at a closing speed of 30 mph. The time required is:

\[
\ ext{Time} = \frac{\ ext{Distance}}{\ ext{Relative Speed}} = \frac{120 \, \ ext{miles}}{30 \, \ ext{mph}} = 4 \, \ ext{hours}
\]

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Total Time Since Departure

Since Train B starts 2 hours after Train A, and takes 4 hours to catch up, the total time after Train A’s departure is:

\[
2 \, \ ext{hours} + 4 \, \ ext{hours} = 6 \, \ ext{hours}
\]