Two bikers simultaneously starting driving towards each other from their homes. They pass each other, reach the other's house, and drive back, all at constant speed. Up to their second meeting, biker Y drives 60 miles more than his friend. Biker B gets home $4\frac 23$ hours after their second meeting. Biker Y gets home $\frac 25$ hours after their second meeting.

I need to find the distance between their houses, and the speed of rider Y.

I wrote $x_B(t)=v_Bt,x_Y(t)=v_Yt$. I wrote equations $60+x_B(t_2)=x_Y(t_2)$ and $x_B(t_2+4\frac 23)=x_Y(t_2+\frac 25)$. By subtracting I found $60-v_B4\frac 23=-\frac 25 v_Y$, but this doesn't get me anywhere. I think I didn't get everything from the wording but I'm stuck.


You have only found two equations in three unknowns, $v_B, v_Y, t_2$. You need to add in the fact that the total distance they cover after $t_2$ adds up to the distance between the houses. How many times have they covered the distance between the houses up to $t_2$?

  • $\begingroup$ Three times in total. Thank you $\endgroup$ – user372838 Sep 27 '16 at 14:37

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