Explanation
To solve the question, we will have to make use of the fact that
[tex]time=\frac{distance}{speed}[/tex]So, we can make the initial speed = v
Thus, we will have the equation:
For the first part
[tex]time=\frac{14}{v}[/tex]For the second portion, since he reduced the speed by 2miles/hour, the time will be
[tex]time=\frac{10}{v-2}[/tex]Also, we are told the time for the first and second part of the journey are equal, therefore
[tex]\frac{14}{v}=\frac{10}{v-2}[/tex]If we solve for v
[tex]\begin{gathered} 14(v-2)=10v \\ 14v-28=10v \\ 14v-10v=28 \\ 4v=28 \\ \\ v=\frac{28}{4} \\ \\ v=7 \end{gathered}[/tex]Therefore, the cyclist's speed during the first portion is 7 miles per hour
Also
His speed during the cooldown portion will be (7-2)miles per hour = 5 miles per hour
