Two trains travel toward each other on the same track, beginning 100 miles apart. One train travels at 40 miles per hour; the other travels at 60 miles an hour. A bird starts flight at the same location as the faster train, flying at a speed of 90 miles per hour. When it reaches the slower train, it turns around, flying the other direction at the same speed. When it reaches the faster train again, it turns around -- and so on. When the trains collide, how far will the bird have flown?
Bird
(Faster Train) A---------------------------------------B (Slower train)
My Attempt:
x: Time at which the two trains collide
100-40x=60x ====> x=1hr
x: Time at which the bird and train B collide
100-40x = 90x ====> x= 10/13 hr
90(10/13) - 90 ( 3/13) = 48 miles
Is this the right approach to the problem? Thanks in advance
Just another version of solving this type of problem.
The Fly and the Trains
A Golden Oldie
Consider the familiar problem of the bouncing ball where a ball is dropped from some height, say one yard high. The ball bounces to 1/2 of its original height and falls again. It again bounces to 1/2 of its previous height and falls again. Theoretically, it will continue to bounce forever but we all know that it actually doesn't. The question is how much linear distance did the ball cover in falling from 3 feet high until it came to rest?
Lets examine what the process produces in the way of real distances. The first drop results in the ball traveling 1 yard. The first bounce results in the ball traveling up 1/2 yard. The second fall results in the ball traveling another 1/2 yard. The second bounce results in the ball traveling up 1/4 yard. The next fall results in the ball traveling another 1/4 yard. If we continue this process, what do we derive? Lets examine the series of distance traveled.
Fall...........1........1/2........1/4........1/8........1/16........1/32.............
Bounce........1/2........1/4........1/8........1/16........1/32...................
If we pair up the equal values, we end up with the sequence of round trips of the ball as
1 + 1 + 1/2 + 1/4 + 1/8 + 1/16 + 1/32 +...............
Expressed another way, 1 + 1 + (1/2)^1 + (1/2)^2 + (1/2)^3 + (1/2)^4 + ............or
1 + [1 + (1/2)^1 + (1/2)^2 + (1/2)^3 + (1/2)^4 + ............(1/2)^n]
The sequence inside the parenthesis is a geometric progression of the form
[1 + r + r2 + r^3 + r^4 + ............r^n] where -1 , r , +1.
The limiting sum of this infinite sequence is given by S = 1/(1 - r).
Therefore, the distance that our ball travels in total is D = 1 + 1/(1 - 1/2) = 1 + 2 = 3 yards.
If you think about it, our trains and fly problem appears to be a similar problem in that with a given set of parameters, the fly will travel some percentage of the starting distance between the two trains before meeting up with the oncoming train. The two trains will now be some smaller distance apart. I suspect that the distance traveled on the return trip will be the same percentage of this smaller distance between the trains. This could lead to a similar geometric series that might lead us to an answer. Lets examine a much simpler case to see what develops.
Consider the folowing scenario:
...............I............................................................................I
.............AI---->Va........ .............D............................Vb<----IB
.............FI---->Vf.................................................................I
The trains are designated by A and B. Va and Vb are the speeds of the trains. Vf is the speed of the fly. D is the starting distance between the trains when the fly begins his journey.
For initial simplicity, assume Va = Vb = Vf = V. since both the train and fly are traveling at the same speed, they will simultaneously meet up with train B in time t. The two trains (and the fly) are approaching one another at the combined speed of Vf + Vb = 2V .The fly reaches train B in t = D/(2V). The fly travels D(V)/(2V) = D/2 miles in that time. Rather obvious.
Now assume that Vf = Va + Vb = 2V. The fly reaches train B in t = D/(3V). The fly travels DVf/(3V) = D(2V)/(3V) = 2D/3. During that same time t, train A travels DV/(3V) = D/3. Thus, the distance between the two trains is now 2D/3 - D/3 = D/3. The ratio of the distance the fly travels to the starting distance between the trains is (2D/3)/ D = 2/3.
On the return trip, the fly reaches train A in t = (D/3)/(3V) = D/(9V). The fly travels D(2V)/(9V) = 2D/9 miles in that time. During that same time t, train B travels DV/(9V) = D/9. Thee distance between the two trains is now D/3 - 2D/9 = D/9. The ratio of the distance the fly travels to the starting distance betwwen the two trains is (2D/9)/(D/3) = 2/3. What have we here?
If we continue the process, we would find that the fly travels 2/3rds of the starting distance between the two trains in each cycle. The sum of these distances can be express by
S = 2D/3 + 2D/9 + 2D/27 + 2D/81 + ......2D/3^n or
S = 2D[1/3 + 1/9 + 1/27 + 1/81 + ...........1/3^n] or
S = 2D/3[1 + 1/3 + 1/9 + 1/27 + 1/81 + .......1/3^n] or
S = 2D/3[1 + (1/3)1 + (1/3)^2 + (1/3)^3 + .........(1/3)^n] or
S = 2D/3[1/(1 - (1/3))] = 2D/3[1/(2/3)] = 2D/3(3/2) = D.
Therefore, if the two train speeds are equal and the speed of the fly is the sum of the two train speeds, the distance the fly travels is equal to the starting distance between the two trains.
What if the two train speeds are not the same and the speed of the fly is still the sum of the two train speeds. (Note that the distance the fly travels on each leg is equal to the sum of the distances traveled by the two trains since the speed of the fly is equal to the sum of the two train speeds.) Without going through the sequence of steps, each leg of one round trip of the fly results in two different ratios of distances traveled to starting distances between trains. However, the sum of the two distances traveled by the fly in one round trip divided by the starting distance at the beginning of each round trip is a constant. What this means is that we can sum the distances of each round trip much the same as we summed the individual leg distances.
If we let the ratio of distance traveled by the fly to the starting distance of each round trip equal p/q, the distance traveled during the first round trip becomes D(p/q). The starting distance between the two trains for the second round trip then becomes D - D(p/q) = D(1 - pq).
Starting the second round trip, the fly travels (p/q)D(1 - p/q).The starting distance for the third round trip then becomes D(1 - p/q) - D(1 - p/q)(p/q) = D(1 - p/q)^2. The distance traveled by the fly during the next round trip becomes (p/q)D(1 - p/q)^2. This leads to the sum of the distances traveled being
S = D(p/q) + D(p/q)(1 - p/q) + D(p/q)(1 - p/q)^2 + D(p/q)(1 - p/q)^3 + ...............
S = D(p/q)[1 + (1 - p/q) + (1 - p/q)^2 + (1 - p/q)^3.................
S = D(p/q)[1/(1 - p/q)] = D(p/q)(q/p) = D.
Therefore, given two different train speeds and the fly speed equal to the sum of the two train speeds, the distance traveled by the fly is equal to the starting distance between the two trains.
The obvious next question is what if the speed of the fly is not equal to the sum of the two train speeds? Fear not, under these circumstances, the distance traveled by the fly becomes
........................................S = DVf/(Va + Vb)......I kid you not.
Getting back to our original problem, S = 200(80)/(40 + 60) = 8000/100 = 160 miles.
Now that we have learned the long way, I'll let you in on the method that gives you the answer almost instantaneously.
With train speeds of 40 and 60mph, the two trains will be approaching one another at the closing speed of 100mph. Therefore, starting out 200 miles apart, they will meet in 200/100 = 2 hours. Since the fly is traveling at a constant speed of 80mph all that time, his total distance traveled must be S = 80(2) = 160 miles.
Don't be upset. Knowing the logical approach and the analytical approach to this oldie should alert you to seek different approaches to problems you will encounter in the future.
