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Two trains leaving the station at different times...

Two trains are driving toward one another. The first train leaves Town A at 5am traveling at 60 miles per hour. The second train leaves Town B at 7am traveling at 70 miles per hour. the distance between Town A and Town B is 455 miles. What is the EXACT time that the collision will occur?

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Gene G. | You can do it! I'll show you how.You can do it! I'll show you how.
5.0 5.0 (257 lesson ratings) (257)

The trick with word problems is to pick them apart to see what you know, then build equations that you can solve.  It takes some practice.

In this problem, we know that one train starts two hours before the other one and travels at a known rate. Let's calculate how far it goes in those two hours:

d = rt = 2 hrs * 60 mi/hr = 120 mi
The time is now 7:00 and the trains are 455 - 120 = 335 mi apart.

Now both trains are moving, one traveling at 60 mi/hr and the other at 70 mi/hr.
Their closing speed is just 60 + 70 = 130 mi/hr, and the total distance to go is 335 mi.

Solve the rate equation for t to get t = d/r.  Plug in 335 for d and 130 for r.
Time to collision is t = 325 / 130 = 2.5769 hrs.

That's 2 hours with a remainder of 0.5769
0.5769 * 60 minutes = 34.615 minutes
0.615 * 60 seconds = 37 seconds

The time of collision will thus be the sum of 2 hrs + 2.5769 hrs added to the 5:00 starting time.
Tc = 5:00 + 4.5769 = 5:00 + 4:00 + 0:34 + 0:00:37 = 9:34:37

Sometimes, a problem will even ask where they would meet.  Once you get to the point where you know the total travel times of both trains and their travel times, you could calculate how far each one went and find the exact location where they meet.  You can treat each train's distance as a separate problem since you now know what each one did independently of the other.  Try it!  To check, their distances added together should be 455 when you're finished.

I hope this is helpful.

Brad M. | STEM Specialist plus Business, Accounting, Investment & EditingSTEM Specialist plus Business, Accountin...
4.9 4.9 (233 lesson ratings) (233)

Hey Evan -- here's the "play by play" ... Train A goes 120mi from 5-7am, leaving a 335mi gap ... after 7am, the gap closes at 130mph ... "stair-step" ... 8am, 205mi left ... 9am, 75mi remains ... 9:30am, 10mi left -- need 1/13 more of an hour ... 60min/13 = 4.6 mins ==> crash at 9:34:35 :)

Patricia S. | Math Tutoring for K-12 & CollegeMath Tutoring for K-12 & College
5.0 5.0 (39 lesson ratings) (39)

Hi, Evan.

Word problems can be tough.  The best way that I've found to approach this type of question is by drawing a picture.  Here's mine:

      [Train #1]  -->                      *Boom*                  <--  [Train #2]

(Town A = Mile #0)                                                  (Town B = Mile #455)

So if Train #1 is traveling at 60 miles per hour, the distance that this train travels can be represented by the equation d = 60x (d is distance, x is the number of hours the train has been traveling).  Seeing as how I chose for Town A to be at mile #0, the equation d = 60x also tells us how far Train #1 is from Town A.

Likewise, Train #2 is traveling at 70 miles per hour, so the distance this train travels can be represented by the equation d = 70(the variables mean the same thing).  However, the equation d = 70x does NOT represent the distance Train #2 is from Town A, because it left from Town B.  Instead, if we write the left side of the equation as 455-d, we are able to find out how far Train #2 is from Town A.  Our new equation for Train #2 is 455-= 70x

The question wants you to figure out when the trains are at the exact same point on the tracks (in my picture above, when the trains go *Boom* :) ).  In other words, when are both trains the same distance away from Town A.  Now, the trains don't leave their stations at exactly the same time, so we can't just replace the d in Train #2's equation with 60x.  Somehow, we have to represent the time difference in their departure in one or both of the equations.  Here's a nice table to demonstrate the issue:


           Train #1       Train #2

5AM         0                 --            <-- Train #1 leaves the station at Town A, but hasn't traveled any miles yet, so it is 0 miles away from Town A.

6AM         60               --

7AM        120               455        <-- Train #2 leaves the station at Town B, but hasn't traveled any miles yet, so it is 455 miles away from Town A.

8AM        180               455-70=385

9AM        240               385-70=315

10AM      300               315-70=240  <-- Somewhere between 9AM and 10AM, our two trains hit each other.

Let's look at our equation for Train #1 again:   = 60x.

When Train #2 begins to travel, Train #1 has already traveled 120 miles away from Town A (See table above.).  If we write our equation for Train #1 as d = 60x + 120, now both this equation and the equation that we have for Train #2 refer to the distance each train is away from Town A, starting at 7AM (the first time that BOTH trains are traveling).

So, to recap:

Train #1 equation:  = 60x + 120 ;      Train #2 equation: 455 - = 70x

Both equations use the same meaning for d and for x, so your next step would be to combine them into one equation and solve for (which should be the only variable left in your combined equation).  Let me know if you need more help or if you have any questions about the explanation thus far.

Good luck!


Danielasdf B. | "English, SPSS, And Biology Tutor""English, SPSS, And Biology Tutor"

The first train travels 60 miles from 6am to 7am
So at 7am the trains are 455 - 60 = 395 miles apart
First train: Distance from town A (DA) = Velocity* time = 60t
Second train: Distance from town B (DB) = Velocity*time = 70t
So, DA + DB = 60t + 70t
395 = 130t
t = 395/130
= 3.4 hrs
= 3 hrs 4 mins
The first train left at 5am so the exact time would be 8:04am (5am +3 hrs and 4 minutes)