Question

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 5km/hr faster than the second train. if after two hours , they are 50 km apart, find the average speed of each train​

Answer 1

Answer:

Let x km /hour be the speed of the first train

Then the speed of the second train is (x - 5) km/hour

Given that they are 50 km apart after 2 hours

As per the given conditions

The first train travels from O to A and

The 2nd train travels from O to B

Also given that AB = 50 Km

Since the OAB is a right angled triangle

∴ By Pythagoras theorem

OA2 + OB2 = AB2………….1

Now distance = speed × time

⇒ OA = x × 2

Also OB = 2(x - 5)

Putting value of OB and OA in 1

⇒ 8 x2 – 40x + 100 = 2500

⇒ x2 – 5x – 300 = 0

⇒ x(x - 20) + 15(x - 20) = 0

⇒ (x + 15) (x - 20) = 0

⇒ x = - 15 or 20

Since speed cannot be negative ∴ the speed of the 1st train = 20 km/hour

And speed of 2nd train is 15 km/hour

Answer 2

Answer:

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Step-by-step explanation:

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 equation4s^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

:

:

Check this; find the distance (d) between the trains using these distances

Northbound traveled 2(15) = 30 mi

Westbound traveled 2(20) = 40 mi

d =

d = 50