Assuming that both cyclists are traveling AWAY from each other with the same starting point, consider the following:
Let us denote the speed of the first bicyclist by [tex]s[/tex] (measured in miles per hour). Then, the speed of the second bicyclist must be [tex]s-5[/tex] (miles per hour).
To verify this, assign an arbitrary number, say 17 mph, to the first bicyclist. Since the first bicyclist is always 5 mph faster than the second, we must have 17 - 5 = 12 mph as the second speed.
At the 0th hour,
- first bicyclist has traveled 0 miles
- second bicyclist has traveled 0 miles
- distance between both = 0 miles
At the 1st hour,
- first bicyclist has traveled [tex]1\times s=s[/tex] miles
- second bicyclist has traveled [tex]-1\times(s-5)=-s+5[/tex] miles (note the negative sign here. Since they are traveling in opposite directions, they must have opposite signs of each other in distances traveled)
- distance between both = [tex]s-(-s+5)[/tex] = [tex]s+s-5[/tex] = [tex]2s-5[/tex] miles
At the second hour,
- first bicyclist has traveled [tex]2s[/tex] miles
- second bicyclist has traveled [tex]-2(s-5)=-2s+10[/tex] miles
- distance between both = [tex]2s-(-2s+10)=2s+2s-10=4s-10[/tex] miles
According to the question statement, both of them are 70 miles apart at the second hour, i.e.
[tex]4s-10=70[/tex]
such that solving this gives:
[tex]4s=70+10[/tex]
[tex]4s=80[/tex]
[tex]s= \frac{80}{4} [/tex]
[tex]s=20[/tex] mph
Therefore, their rates are as follows:
- first bicyclist = 20 mph
- second bicyclist = 20 - 5 = 15 mph
