Question

Two cyclists start at the same point and travel in opposite directions. One cyclist travels 7 mi/h slower than the other. If the two cyclists are 123 miles apart after 3 hours, what is the rate of each cyclist?

Answer 1

The speed of faster cyclist is 85.6 m/h

The speed of slower cyclist is 78.6 m/h

Explanation:

Given as :

The total distance between two cyclist = d = 123 miles

The Time taken by cyclist to move apart = t = 3 hours

Let The speed of faster cyclist = s miles per hour

And The speed of slower cyclist = (s - 7) miles per hour

Now, According to question

Time =
`(\textrm distance)/(\textrm speed)`

So, 3 =
`(123)/(s)` +
`(123)/(s - 7)`

Or,
`(3)/(123)` =
`(1)/(s)` +
`(1)/(s-7)`

Or,
`(1)/(41)` =
`(s + (s - 7))/(s(s -7))`

Or, s(s - 7) = 41 (2 s - 7)

Or, s² - 7 s = 82 s - 287

Or, s² - 7 s - 82 s + 287 = 0

Or, s² - 89 s + 287 = 0

Now, Solving this quadratic equation

s =
`\frac{- b\pm \sqrt{b^(2) - 4* a* c}}{2* a}`

Or, s =
`\frac{- (-89)\pm \sqrt{(-89)^(2) - 4* 1* 287}}{2* 1}`

Or, s =
`(89+82.2)/(2)` ,
`(89-82.2)/(2)`

∴ speed = 85.6 m/h , 3.4 m/h

Here we consider 85.6 m/h

So, The speed of faster cyclist = s = 85.6 m/h

And The speed of slower cyclist = s - 7 = 85.6 - 7 = 78.6 m/h

Hence,The speed of faster cyclist is 85.6 m/h

And The speed of slower cyclist is 78.6 m/h Answer