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?

## Two trains leaving the station at different times...

# 4 Answers

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.

T_{c} = 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.

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 :)

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* = 60*x* (*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* = 60*x* 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*x *(the variables mean the same thing). However, the equation
*d* = 70*x* 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-*d *= 70*x*.

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:

HOW FAR AWAY EACH TRAIN IS FROM TOWN A

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: *d *= 60*x.*

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

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 *x *(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!

~Patty

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)