Question

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

Answer 1

Let the speed of second train be xkm/hr So ,the speed of first train will be (x+5 )km/hr / |N / | 50 / |y / __ |____E W | M distance travelled By - 1st train = speed×time x(2) 2x = M 2nd train - (x+5)×2 2(x+5)=y Now by Pythagoras theorem (50)^2 = M^2 +y^2 2500 = 2x^2 + {2(5+x)}^2 2500 = 4x^2 + 4(25 +x^2 +10x) 2500 = 4x^2 + 100 + 4x^2 + 40x 2500 -100 = 8x^2 + 40x 2400 = 8x^2 + 40x 0 = 8x^2 + 40x - 2400 { talking 8 common} 0 = x^2 + 5x - 300 0= x^2 +20x -15x - 300 0= x(x +20) -15(x + 20) 0 = (x-15) (x+20) x - 15 =0 x =15 And x +20 =0 x = -20 speed (x) cant be -ve Thus ,x = 15 Hence, the speed of 1st train is 20kmphr and 2nd train is 15kmphr

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