Question

Two cyclists, 108 miles apart, start riding toward each other at the same time. One cycles 2 times asfast as the other. If they meet 4 hours later, what is the speed (in mi/h) of the faster cyclist?

Answer 1

Initial distance: 108 miles

We know that they start riding toward each other, and one of them is 2 times as fast as the other. Then, if the speed of the slowest is v, the speed of the faster cyclist is 2v. The combined speed is:

`v_T=v+2v=3v`

The speed and the distance are related by the equation:

`V=(D)/(t)`

They meet 4 hours later, thus:

`\begin{gathered} D=108 \\ t=4 \end{gathered}`

Finally, using the previous equation:

`\begin{gathered} 3v=(108)/(4) \\ \Rightarrow v=9\text{ mi/h} \end{gathered}`

The speed of the faster cyclist (2v) is 18 mi/h.