Question

$\huge \sf \color{purple}{Question}$


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

Answer 1

$\huge \fbox \red{Answer}$

[Refer the attachment for help]

Let speed of the second train be x km/hr.

Then, the speed of the first train is (x + 5) km/hr.

Let O be the position of the railway station from which the two trains leave.

Distance traveled by the first train in

2 hours = OA

= Speed × Time

= 2(x + 5) km

Distance traveled by the second train in

2 hours = OB

= Speed Time

= 2x km

By using Pythagoras Theorem, we have,

AB² = OA² + OB²

50² = {2(x + 5)}² + (2x)²

2500 = 4(x + 5)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

x = -20 or, x = 15

x = 15 [because cannot be negative]

Hence, the speed of the second train is 15 km/hr and, the speed of the first train is 20 km/hr.

Answer 2

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

Speed of first train is 20 Km/h and second 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