A and B are two stations 300 miles apart. 2 trains start simultaneously from A and B, to the opposite station. Find the each train's speed if...

Question

The train from A reaches B 9 hours after the 2 trains meet, and the train from B reaches A 4 hours after they meet.

Answer 1

What a neat question!

 

I found the solution. The difficult part was organizing the given information in a way that led to a solution. Typing out the algebra is not going to look very pretty so be warned! :)

SETUP INFORMATION
The problem requires us to consider the motion of the trains before they meet AND after they meet. So, I let the meeting point be x miles from station A. Since station B is 300 miles away, the meeting point is also 300-x miles from station B.

For clarity, I will refer to train that leaves from station A as train A. Likewise, the train that leaves from station B is train B.

Also, we will be using the equation s = d/t, where s is the speed, d is the distance, and t is the time. Lastly, let the time that the trains meet be represented by a capital T (in hours).


The speed of each train does not change during the trip so a train’s speed before they meet is the same as its speed after they meet.

TRAIN A
The speed before they meet is the given by the distance traveled before they meet divided by the time it takes to meet. This means the speed of train A is x/T.

The speed of train A after they meet is given by the distance traveled after they meet divided by the time it takes to reach station B. This means that the speed can also be written as (300-x)/9.

So, x/T = (300-x)/9. (Let’s call this Equation 1 and save it for later)


TRAIN B
Let’s repeat this process for train B. The speed before they meet is the given by the distance traveled before they meet divided by the time it takes to meet. This means the speed of train B is (300-x)/T.

The speed of train B after they meet is given by the distance traveled after they meet divided by the time it takes to reach station B. This means that the speed can also be written as x/4.

So, (300-x)/T = x/4. (Let’s call this Equation 2)
If we use this equation, and cross multiply, we can solve for x (its going to look ugly!!)

4(300-x) = xT
1200-4x = xT
1200 = xT+4x
1200 = x(T+4)
1200/(T+4) = x (Call this Equation 3)

Now, lets revisit Equation 1 and cross multiply

x/T = (300-x)/9
9x = T(300-x)
9x = 300T – xT
9x + xT = 300T
x(9+T) = 300T
x = 300T/(9+T) (Call this Equation 4)

Notice that Equation 3 and Equation 4 both tell us what x equals. Since, x can’t be two different numbers, they must be equal. So,

1200/(T+4) = 300T/(9+T) Now, cross multiply
1200(9+T) = 300T(T+4)
10800 + 1200T = 300T^2 +1200T

Subtract 1200T from both sides, and

10800 = 300T^2
36 = T^2
6 = T

The trains meet in 6 HOURS.

Since train A takes 9 more hours to reach station B, its travel time is 15 hours.
300 miles in 15 hours is 20 miles/hour.

Since train B takes 4 more hours to reach station A, its travel time is 10 hours.
300 miles in 10 hours is 30 miles/hour.


A quick check shows that at 20 miles/hour, train A travels 120 miles in 6 hours and train B, at 30 miles/hour, travels 180 miles in 6 hours. The 120 added to the 180 equals 300 which is the distance between the stations!

Hope this makes sense!! :)

Answer 2

Assuming that the train reaches the other station in the time given, you only need to set up two equations.

speed = miles / hour  

(knowing the units helps with putting the correct values on top and bottom of the equation)

Train A : 300 miles/ 9 hours= 33.3 mph

Train B : 300 miles/ 4 hours= 75 mph

 

However, as Jim pointed out, this is not the entire time that the train is traveling. Therefore, the equation instead becomes Speed of train A (Sa) = 300 / (t+9) Speed of train B (Sb) = 300 / (t+4) Where t is the time they meet The trains meet at some point x, which I made closer to a since it is traveling slower Sa= x/t we don't know how long it takes for Train A to get here Sb= x/4 but we do know that it took Train B 4 hours to get to station A so we can set these two equations equal Sa = x/t = 300 / (t+9) Sb = x/4 = 300 / (t+4) Then solve for x and set them equal (because we don't need to know the distance where they meet) x= 300 t / (t+9) = 300 *4 / (t+4) the 300 cancels out and multiply the denominators on each side t (t+4) = 4 (t+9) move everything to one side and distribute t^2 + 4t - 4t -36 = 0 combine like terms t^2 - 36 = 0 factor (t +6) (t-6 ) = 0 time cannot be negative so ignore that answer t=6 now plug into the original to find speed Sa= 300 / (6+9) = 20 miles per hour Sb= 300 / (6+4) = 30 miles per hour