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MariettaO [177]
3 years ago
8

Two trains travel at right angles to each other after leaving the same train station at the same time. Two hours later they are

70.46 miles apart. If one travels 9 miles per hour slower than the other, what is the rate of the faster train? (Round your answer to the nearest integer)
Physics
2 answers:
trasher [3.6K]3 years ago
6 0

Answer:

The faster train travels at a speed of 29 miles/hr    

Explanation:

Let the speed of train A be x miles/hr

So the speed of the train B is ( x - 9 ) miles/hr

After two hours, their speed will be 2x and 2( x - 9 ) respectively. and the distance between them is 70.46 miles.

So using Pythogorus theorem, we can write

(2x)^{2}+(2x-18)^{2} = 70.46^{2}

8x^{2}-72x+324 = 4964.6116

8x^{2}-72x-4640.6116 = 0

On solving the above equation we get

x = 29.0015 miles/hr

 = 29 miles/hr

Therefore the faster train travels at a speed of 29 miles/hr

Vikentia [17]3 years ago
3 0

Answer:

Speed of faster train equals 29 mph

Explanation:

Let the speed of slower train be 'x' miles per hour and the speed of faster train be 'y' miles per hour.

Distance that slower train covers in 2 hours=2\times x miles

Distance that faster train covers in 2 hours=2\times y miles

Since they move at right angles the distance between them can be found by Pythagoras formula as

d^{2}=(2x)^{2}+(2y)^{2}\\\\d^{2}=4(x^{2}+y^{2})\\(70.46)^{2}=4(x^{2}+y^{2})\\\\\therefore (x^{2}+y^{2})=\frac{(70.46)^{2}}{4}\\\\(x^{2}+y^{2})=\frac{4964.6}{4}\\\\(x^{2}+y^{2})=1241.15

It is given that y=x+9

Using this in the above equation we get

(x^{2}+y^{2})=1241.15\\\\x^{2}+(x+8)^{2}=1241.15\\2x^{2}+16x+64=1241.15\\2x^{2}+16x-1177.15=0

This is a quadratic equation in 'x'

Comparing with standard quadratic equation ax^{2}+bx+c we get value of x as

(x^{2}+y^{2})=1241.15\\\\x^{2}+(x+8)^{2}=1241.15\\2x^{2}+16x+64=1241.15\\2x^{2}+16x-1177.15=0\\\\x=\frac{-16\pm \sqrt{(16)^{2}-4\times 2\times -1177.15}}{2\times 2}\\\\x=\frac{-16\pm 98.35}{4}\\\\x=20.58mph(\because speed\neq

Thus speed of faster train = 28.58 mph

Speed of faster train = 19 mph

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