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?

Respuesta :

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:

[tex]v_T=v+2v=3v[/tex]

The speed and the distance are related by the equation:

[tex]V=\frac{D}{t}[/tex]

They meet 4 hours later, thus:

[tex]\begin{gathered} D=108 \\ t=4 \end{gathered}[/tex]

Finally, using the previous equation:

[tex]\begin{gathered} 3v=\frac{108}{4} \\ \Rightarrow v=9\text{ mi/h} \end{gathered}[/tex]

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