Question

Two trains leave a railway station at the same time. The first train travels towards west and the second train towards north. The first train travels 5 km/hr faster than the second train. If after two hours they are 50 km. apart find the average speed of each train.

Answer 1

Distance = speed x time

time = 2 hours

let the speed of train going north = x

speed of train going west = x+5

now d = s*t

so , d for 1st train going north be = 2x

d for 2nd train going west be = 2(x+5)

now this is the case of a right angle triangle so

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

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

2400 = 8x^2 + 40x

dividing LHS AND RHS by 8

300 = x^2 + 5x

x^2 + 5x - 300 = 0

x^2 + 20x - 15x - 300 = 0

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

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

x = 15, -20

ignoring negative value we get,

x = 15

so speed of train 1 going north is 15 kmph

and speed of train 2 going west is 15+5 = 20 kmph

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