2 cyclist start from the same point at the same time. one goes due north and the other due west along straight hihways . also the speed of one cyclist is 5km/hr more than the other . if after 2 hours they are 50km apart from each other find the average speed of each cyclist
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it will form a triagle with hypotenuse = 50 km
and let other sides be a and b
let the speed of one cyclist is x km/h
then speed of second one be x+5 km/h
now, a = speed×time = 2 × x = 2x km
and b = 2(x+5) = 2x + 10 km
applying pythagorus theorem,
(2x)^2 + (2x+10)^2 = (50)^2
4x^2 + 4x^2 +100+ 40x = 2500
8x^2 + 40x - 2400 = 0
x^2 + 5x - 300 = 0
(x-15)(x+20) = 0
x=15 or x=-20
hence, we will take the positive value only because speed can't be negative.
therefore, their average speeds are 15 and 20 km/h
hope, it will help you and don't forget to choose the brainliest.
and let other sides be a and b
let the speed of one cyclist is x km/h
then speed of second one be x+5 km/h
now, a = speed×time = 2 × x = 2x km
and b = 2(x+5) = 2x + 10 km
applying pythagorus theorem,
(2x)^2 + (2x+10)^2 = (50)^2
4x^2 + 4x^2 +100+ 40x = 2500
8x^2 + 40x - 2400 = 0
x^2 + 5x - 300 = 0
(x-15)(x+20) = 0
x=15 or x=-20
hence, we will take the positive value only because speed can't be negative.
therefore, their average speeds are 15 and 20 km/h
hope, it will help you and don't forget to choose the brainliest.
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