I am stuck on this problem that someone gave me. Two cyclists head off simultaneously from two towns situated at a distance of $240$ km length.
Knowing that one covered the same distance in 3 hours(lets denote as the first one) as the other did in 5 hours and the time until they met is 6 hours. Find how much time would it take to the second cyclist to get at the half part of the road (after 6 hours passed of course).
I would really like to find a solution, but I am pretty bad as such types of problems. Denoting velocity=v time=t and d= distance. Here is my try: First I try to find the velocity of each cyclist. $$v=\frac{d}{t}=40\frac{km}{h}$$ This velocity is equal to the average velocity of the cyclists $v=\frac{v_A+v_B}{2}$ Also knowing that $v_A=\frac{5}{3}v_B$ I get that $v=\frac{8}{6}v_B$ which gives the velocity of the driver B is $30\frac{km}{h}$ yielding to $v_A=50\frac{km}{h}$ So the distance done by B is $30\cdot 6=180$km But that is already after half of the road... I would appreciate to get some help in order to get on the half part of the road.
Assume that $3x$ km is the distance the first one travels in $3$ hours or the distance the second one travels in $5$ hours.
$\Rightarrow$ $x$ km/h is the velocity of the first cyclist and $0.6x$ km/h is the velocity of the second cyclist.
They need to cycle for $6$ hours in total before they met, which means they take $6$ hours together to cover the entire distance $240$ km, which means each hour, the total distance covered by both cyclists is $40$ km
$\Rightarrow x+0.6x=40$, is this enough?