Question

Two electric trains leave a railway station at the same time. The first train travels due west and the second train travels due North. The first train travels 5 km/hr faster than the second train. If after an hour they are 25 km apart then speed of the first train is. A) 20 km/hr B) 15 km/hr C) 25 km/hr D) 10 km/hr Please explain step by step ....

Answer 1

Answer:

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

:

:

Answer:

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

:

:

Step-by-step explanation:

Let the speed of the second train = x km/hr

The speed of the first train = x+5 km/hr

Distance covered after two hours by the first train is 2(x+5) km.

Distance covered by the second train after two hours is 2x km.

(2x)^2 + 4(x+5)^2=(50)^2 (Using pythagoras theorem)

Solve

We get x=-20 and x=15

The speed of the first train = 15+5=20km/hr

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Step-by-step explanation:

Let the speed of the second train = x km/hr

The speed of the first train = x+5 km/hr

Distance covered after two hours by the first train is 2(x+5) km.

Distance covered by the second train after two hours is 2x km.

(2x)^2 + 4(x+5)^2=(50)^2 (Using pythagoras theorem)

Solve

We get x=-20 and x=15

The speed of the first train = 15+5=20km/hr

Please follow me dear friend ❤️.

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