SOLUTION: Two cyclists leave the same place at the same time but travel in different directions and at different speeds. Cyclists A travels north at a constant speed off x km/h. Cyclist B tr
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Question 536715: Two cyclists leave the same place at the same time but travel in different directions and at different speeds. Cyclists A travels north at a constant speed off x km/h. Cyclist B travels east at a speed 5km/h more than cyclist A. After 1 hour, the cyclists are 37km apart. At what speed is cyclist A traveling?
Thank you:)
Answer by mananth(16946) (Show Source): You can put this solution on YOUR website!
they are travelling at right angles to each other.
At any given instant they form a right triangle with their starting point
Apply Pythagoras Theorem
speed of DISABLED_event_one= x
other = x+5
X^2 + (x+ 5 }^2 = 37 ^2
X^2 + X^2+ 10 x * 25 = 1369
2 X^2 + 10 x- -1344 = 0
Find the roots of the equation by quadratic formula
a= 2 ,b= 10 c -1344
b^2-4ac= 100 - 10752
b^2-4ac= 10852
= 104.17
x1=( -10 + 104.17 )/ 4
x1= 23.54
x2=( -10 -104.17 ) / 4
x2= -28.543
Ignore negative value
speed = 23.54 mph
speed of cyclist A = 23.54 km/h
Add 5 to get speed of B
m.ananth@hotmail.ca
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