# 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.

• Try using all the information available. Is there some information you haven't used? – Aryabhata Aug 23 '10 at 16:26
• Absolutely no. I'm working on this for like three hours. – Alon Gubkin Aug 23 '10 at 16:46
• Ok, to make the hint more specific: have you used "bike rider['s] speed was 2 times greater than the pedestrian's speed"? Have you used "5 am" and "7 am"? – ShreevatsaR Aug 23 '10 at 16:57
• I edited/rewrote the stated problem to try to make it clearer. Please feel free to revert or re-edit if it does not convey the same meaning. – Isaac Aug 23 '10 at 17:10
• @ShreevatsaR "5 am" and "7 am" - I wasn't using that. Succeed, thanks!!! – Alon Gubkin Aug 23 '10 at 18:14

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.

• If you can't follow any step let me know. – KalEl Aug 23 '10 at 19:11
• Normally it's not a good idea to give such good answers to homework questions, but the questioner claims to have already succeeded, so +1. – ShreevatsaR Aug 23 '10 at 19:25
• Agreed, I wanted to show the way I used think about an algebra problem. While it may not be the best way, it might help him in the future. – KalEl Aug 23 '10 at 19:34

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.