Question

Two trains leave a railway station at the same time. The first train travels towards west and the second tour and travel towards north. First train travels 5 kmph 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 the average spped of the second train be x km/h.

Thus, the speed of the first train will be (x+5) km/h.

Distance travelled by the first train in two hours =2(x+5) km/h

Distance travelled by the second train in two hours =2x km/h

Using pythagoras theorem, we have

4x

2

+4(x+5)

2

=50

2

4x

2

+4x

2

+40x+100=2500

8x

2

+40x−2400=0

x

2

+5x−300=0

(x−15)(x+20)=0

x=15,−20

Ignore the negative value of the speed. We have,

Speed of the second train be 15 km/h.

Spped of the first train be (15+5)= 20 km/h.

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Answer 2

$\huge \sf \color{purple}{\underline{\underline{Answer}}} :$

Speed of first train is 20 Km/h and speed of 2nd train is 15 Km/h

$\huge \sf \color{purple}{\underline{\underline{Solution}}}:$

➣ Let the 2nd train travel at X km/h

➣Then, the speed of a train is (5 +x) Km/hour.

➣ let the two trains live from station M.

➣ Distance travelled by first train in 2 hours

$\sf \boxed{\colorbox{skyblue}{Distance=speed×time}}$

= MA = 2(x+5) Km.

➣ Distance travelled by second train in 2 hours

= MB = 2x Km

$\sf \color{brown}{By \: Phythagoras \: theorem }$ AB²= MB²+MA²

⟹ 50²=(2(x+5)²+(2x)²

⟹ 2500 = (2x+10)² + 4x²

⟹8x² + 40x - 2400 = 0

⟹x² + 5x - 300 = 0

⟹x² + 20x -15x - 300 = 0

⟹x(x + 20) - 15(x + 20) = 0

⟹ (x + 20)(x -15) = 0

$\sf \boxed{\colorbox{lightgreen}{x=15 \: or \: -20}}$

Taking x = 15 , the speed of second train is 15 Km/h and speed of first train is 20 Km/h