Asked by Faith
Two buses leave a city at the same time,one going north and the other east.The north bound bus travels 10km/hr faster than the east bound one.If they are 250km apart after 5hours,What is the speed of each bus?
Answers
Answered by
Bosnian
If one going north and the other east distance will be:
√ { ( 5 x )² + [ 5 ( x + 10 ) ]² } = 250
5 ( x + 10 ) = 5 x + 50
So:
√ [ ( 5 x )² + ( 5 x + 50 )² ] = 250
√ [ 25 x ² + 25 x² + 500 x + 50² ] = 250
√ ( 50 x² + 500 x + 2 500 ) = 250
50 x² + 500 x + 2 500 = 250²
50 x² + 500 x + 2 500 = 62 500
50 x² + 500 x + 2 500 - 62 500 = 0
50 x² + 500 x - 60 000 = 0
The solutions are:
x = - 40 an x = 30
Speed can't be negative, so speed of slower bus:
x = 30 km / h
speed of faster bus = x + 10 = 30 + 10 = 40 km / h
After 5 hrs, the slower went 30 ∙ 5 = 150 km
the faster went 40 ∙ 5 = 200 km
√ [ ( 150)² + ( 200)² ] = √ ( 22 500 + 40 000) = √ 62 500 = 250 km
√ { ( 5 x )² + [ 5 ( x + 10 ) ]² } = 250
5 ( x + 10 ) = 5 x + 50
So:
√ [ ( 5 x )² + ( 5 x + 50 )² ] = 250
√ [ 25 x ² + 25 x² + 500 x + 50² ] = 250
√ ( 50 x² + 500 x + 2 500 ) = 250
50 x² + 500 x + 2 500 = 250²
50 x² + 500 x + 2 500 = 62 500
50 x² + 500 x + 2 500 - 62 500 = 0
50 x² + 500 x - 60 000 = 0
The solutions are:
x = - 40 an x = 30
Speed can't be negative, so speed of slower bus:
x = 30 km / h
speed of faster bus = x + 10 = 30 + 10 = 40 km / h
After 5 hrs, the slower went 30 ∙ 5 = 150 km
the faster went 40 ∙ 5 = 200 km
√ [ ( 150)² + ( 200)² ] = √ ( 22 500 + 40 000) = √ 62 500 = 250 km
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