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Question

From a station, two trains start at a same time. One train moves in West and other to North direction. First train moves 5 km/h faster than the second train. If after 2 hours distance between two trains is 50 kms , find the average speed of each train

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Solution

Two trains leave the railway station at the same time.

The first travels due west and second train due north.

The speed of the first train is 5 km faster than the second train.

After two hours they are 50 km apart. Find the average speed of the trains

:

Let s = the speed of the northbound train

Then

(s+5) = the speed of the westbound train

:

This is a right triangle problem: a^2 + b^2 = c^2

The distance between the trains is the hypotenuse

dist = speed * time

The time is 2 hrs, so we have

a = 2s; northbound train distance

b = 2(s+5) = (2s+10); westbound distance

c = 50; distance between the two trains

:

(2s)^2 + (2s+10)^2 = 50^2

4s^2 + 4s^2 + 40s + 100 = 2500

:

Arrange as a quadratic equation

4s^2 + 4s^2 + 40s + 100 - 2500 = 0

8s^2 + 40s - 2400 = 0

:

Simplify, divide by 8:

s^2 + 5s - 300 = 0

:

Factors to

(s - 15)(s + 20) = 0

:

The positive solution is what we want here

s = 15 mph is the speed of the northbound train

then

5 + 15 = 20 mph is the speed of the westbound train

The first travels due west and second train due north.

The speed of the first train is 5 km faster than the second train.

After two hours they are 50 km apart. Find the average speed of the trains

:

Let s = the speed of the northbound train

Then

(s+5) = the speed of the westbound train

:

This is a right triangle problem: a^2 + b^2 = c^2

The distance between the trains is the hypotenuse

dist = speed * time

The time is 2 hrs, so we have

a = 2s; northbound train distance

b = 2(s+5) = (2s+10); westbound distance

c = 50; distance between the two trains

:

(2s)^2 + (2s+10)^2 = 50^2

4s^2 + 4s^2 + 40s + 100 = 2500

:

Arrange as a quadratic equation

4s^2 + 4s^2 + 40s + 100 - 2500 = 0

8s^2 + 40s - 2400 = 0

:

Simplify, divide by 8:

s^2 + 5s - 300 = 0

:

Factors to

(s - 15)(s + 20) = 0

:

The positive solution is what we want here

s = 15 mph is the speed of the northbound train

then

5 + 15 = 20 mph is the speed of the westbound train

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