Traffic Homework Help > At 5am, a pedestrian starts walking from $A$ to $B$, a distance of $30$ km. At $7$ am a bicyclist whose speed is $2$ times the pedestrian's speed, started riding from $A$ to $B$, too. After some time, the bicyclist met and passed the pedestrian. Two hours after the meeting **the pedestrian** reached his destination ($B$). (The bicyclist reached $B$ before the pedestrian reached it.) > > What is the speed of the pedestrian? So, first I found the distance that the pedestrian went until the bike rider reached him like that: * Marked $x$ = the pedestrian's speed. * Marked $y$ = the distance that the pedestrian went until the bike rider reached him. * Time until the bike rider reached the pedestrian = $\frac{y}{x}$. * $x\frac{y}{x + 2} = 30$ * $y = 30 - 2x$ So the distance that the pedestrian went until the bike rider reached him was $30 - 2x$. But how I continue from here? Thanks.

> At 5am, a pedestrian starts walking from A to B, a distance of 30 km.

Let speed of pedestrian be $v$. At time $t$, pedestrian is at distance $vt$. (Where $t$ denotes time since pedestrian started.)

> At 7 am a bicyclist whose speed is 2 times the pedestrian's speed, started riding from A to B, too.

Speed of cyclist is $2v$. At time $t$, cyclist is at distance $(2v)(t-2)$. (Why?)

> After some time, the bicyclist met and passed the pedestrian.

Let they meet at time $t_1$. Then $vt_1=2v(t_1-2)$. (Why? Notice that $v$ cancels out, hence solve for $t_1$ from here.)

> Two hours after the meeting the pedestrian reached his destination.

The pedestrian covers 30km in $t_1+2$ hours, with $t_1$ known. Hence deduce his speed.