# Word Problem

• May 9th 2009, 03:49 PM
Mr Rayon
Word Problem
Alex and Nat are going for a bike ride. Nat can ride at 10km/h, while Alex can develop a maximum speed of 12km/h if he needs to. Nat leaves home at 10 am, while Alex stays behind for 15 minutes and then sets out to catch up with Nat. When will Alex reach Nat, assuming that both of them are riding at their maximum speed?
• May 9th 2009, 04:09 PM
o&apartyrock
15 minutes is on quarter of an hour, so if Nat rides at 10k/h, he will have ridden 2.5 km at the time Alex leaves. Alex rides 2k/h faster than Nat, so he will make up the 2.5 hr different in 1.25 hours, or 1 hour and 15 minutes. since Alex left at 10:15, they will meet at 11:30.
• May 9th 2009, 04:40 PM
Mr Rayon
Quote:

Originally Posted by o&apartyrock
Alex rides 2k/h faster than Nat, so he will make up the 2.5 hr different in 1.25 hours, or 1 hour and 15 minutes.

How did you find out that Alex will make up the 2.5 km difference in 1.25 hours?
• May 9th 2009, 05:07 PM
TheEmptySet
Quote:

Originally Posted by Mr Rayon
Alex and Nat are going for a bike ride. Nat can ride at 10km/h, while Alex can develop a maximum speed of 12km/h if he needs to. Nat leaves home at 10 am, while Alex stays behind for 15 minutes and then sets out to catch up with Nat. When will Alex reach Nat, assuming that both of them are riding at their maximum speed?

Let's set this up with two equations

Nat has a 15 min (1/4) hr head start. So when alex leaves Nat is

$\displaystyle \frac{10km}{h}\cdot \frac{1}{4}hr=2.5km$

ahead. So the equation of his motion is

$\displaystyle d=10t+2.5$

So Alex equation is

$\displaystyle d=12t$

So solving this system of equations gives

$\displaystyle 12t=10t+2.5 \iff 2t=2.5 \iff t=1.25$

So it takes 1.25 hours to catch up.

So the time is 11:30.
• May 9th 2009, 06:58 PM
Mr Rayon
Are these word problems supposed to be easy and straight forward? I find them a bit tricky...