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.
 A: In problems like this (where there are multiple moving objects that start at different times), I usually start by assigning a variable to the quantity that the question asks for: $$\text{let }x=\text{the speed of the pedestrian}$$
then assigning a variable to elapsed time from a specific instant: $$\text{let }t=\text{time (in hours) since 5am}.$$
With these variables, I then try to write expressions for the positions of each moving object in terms of $x$ and $t$, and set the position expressions equal corresponding to when the objects are in the same place.
If this is not enough to get you unstuck, please comment on this answer and let me know what you've done so far using these ideas.
A: 
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.
