Given that the train crosses a 1-mile bridge in 1 minute, we can determine the train’s speed:
1 mile/1 minute = 1 mile/minute = 60 miles/hour.
Now, if the train decelerates to a stop at the midpoint of the bridge, and then accelerates back to its original speed, reaching it by the time it gets to the end, we can model this as two symmetric phases: deceleration and acceleration. We’re looking for the total time spent in these two phases, taking into account the 10 seconds of being stationary in the middle.
Let’s call the time it takes for the first half "T1" and the time it takes for the second half "T2". Due to symmetry, T1 = T2, and we’ll just refer to this time as "T".
Even though the speed is changing, because it’s changing linearly (decreasing to 0, then increasing back to the original speed), the average speed over each half of the journey is just half the original speed: 60/2 = 30 miles/hour.
The distance each half is 0.5 miles, so:
T = 0.5 miles / (30 miles/hour) = 1/60 hours = 1 minute.
So the total time would be 2T (for both halves) plus the 10 seconds the train is stopped in the middle.
So that’s 2 minutes for the movement and 10 seconds for the stop, which gives us a total of:
2 minutes + 10 seconds = 120 seconds + 10 seconds = 130 seconds.
Therefore, it takes the train 130 seconds (or 2 minutes and 10 seconds) to cross the bridge under these conditions.