[lisymathew]

Two trains leave a railway station at the same time. The first train travels due west and the second train
Due north. The first train travels 5 km/hr faster than the second train. If, after 2 hours, they are 50 km
Apart, find the speed of each trains

[Unknown008]

1. Make a quick sketch.
2. Let v be the speed of the slower train, the one going North. The speed of the faster train will then be v+5.
3. Find the distance they travel in terms of v.
4. Using the pythagoras theorem, get the distance they are apart, and equate that to the 50 km. Solve for v.

Can you do that? :)

[smtchahal]

We're given that one train is travelling due west and the second train is travelling due north. This can be seen as a right angled triangle as given below:

[IMG]file:///C:/Users/Sumit/Desktop/Untitled.png[/IMG]

where x refers to the speed of the second train.

Speed of the first train = x + 5 (km/h)
Speed of the second train = x (km/h)

Distance traveled by the first train after 2 hours = 2 (x + 5) = 2x + 10 (km)
Distance traveled by the second train after 2 hours = 2 (x) = 2x (km)

As we can see in the figure, a right-triangle is formed. We can use the Pythagoras Theorem to form a quadratic equation.

By Pythagoras Theorem,

H^2 = P^2 + B^2
=> 50^2 = (2x)^2 + (2x + 10)^2
=> 2500 = 4x^2 + 4x^2 + 100 + 2(2x)(10)
=> 2500 = 8x^2 + 100 + 40x
=> 8x^2 + 40x - 2400 = 0
=> x^2 + 5x - 300 = 0

We have got the quadratic equation => x^2 + 5x - 300 = 0
We can solve it by the quadratic formula
x = [-b ± √(b^2 - 4ac)]/2a
= [-5 ± √(5^2 - 4(1)(-300))]/2(1)
= [-5 ± √(25 + 1200)]/2
= [-5 ± √1225]/2
= [-5 ± 35]/2
Either x = [-5 + 35]/2 = 30/2 = 15
or x = [-5 - 35]/2 = (-40)/2 = - 20

Since speed can't be negative, we'll take x as 15

Therefore the speed of the first train = x + 5 (km/h) = 15 + 5 km/h = 20 km/h
and the speed of the second train = x (km/h) = 15 km/h