Two word problems.

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May 7th 2006, 09:17 AM
Failing Math

Two word problems.

I have to show step by step of how I got BOTH of them.

1.) 2 bikes are 100 miles apart riding at each other at 10mph.

A fly starts at one bicycle and flies towards the other at 60 mph.

When the fly reaches the bike, he turns around and flies back. He continues doing this until they meet.

How much distance did the fly cover?

Total flown by the fly?

PART 2

How much distance did he fly IF...

one bike was going 10 mph and the other was going 20 mph.

2.) On an automobile trip, Aimee Cardella maintained a steady speed for the first two hours. Rush-hour traffic slowed her speed by 25mph for the last part of the trip. The entire trip, a distance of 125 mi, took 2.5 hours. What was her speed during the first part of the trip?
May 7th 2006, 09:22 AM
ThePerfectHacker

Quote:

Originally Posted by Failing Math

I have to show step by step of how I got BOTH of them.

1.) 2 bikes are 100 miles apart riding at each other at 10mph.

A fly starts at one bicycle and flies towards the other at 60 mph.

When the fly reaches the bike, he turns around and flies back. He continues doing this until they meet.

How much distance did the fly cover?

Total flown by the fly?

PART 2

How much distance did he fly IF...

one bike was going 10 mph and the other was going 20 mph.

Let me just be a little of topic.
You can solve this the n00bish way with infinite series. Like von Neumann

Quote:

...I summed the series...

Or the good way.
-----------------------------
The two bicycles are 100 miles. Since each one travels to each other at 20 miles per hour the amount of time required is 5 hours. The fly flies at 60 miles per hour. It flys for 5 hours the total distance it travels is then 300 miles.

The same idea for 10 and 20 miles per hour
May 7th 2006, 09:29 AM
ThePerfectHacker

Quote:

Originally Posted by Failing Math

2.) On an automobile trip, Aimee Cardella maintained a steady speed for the first two hours. Rush-hour traffic slowed her speed by 25mph for the last part of the trip. The entire trip, a distance of 125 mi, took 2.5 hours. What was her speed during the first part of the trip?

Okay.
The distance for the first part of the trip when added to the second part gives the whole distance. Which is 125.
Thus,
$\displaystyle
\mbox{first part }+\mbox{ second part }=125 \mbox{ miles }$
Remember that distance is time multiplied by velocity thus,
$\displaystyle \mbox{first part }=2\times v$
$\displaystyle \mbox{second part }=.5\times 25=12.5\mbox{ miles}$
Thus,
$\displaystyle 2v+12.5=125$
Subtract 12.5 from both sides to get,
$\displaystyle 2v=112.5$
Divide both sides by 2 to get,
$\displaystyle v=56.23 \frac{\mbox{miles}}{\mbox{hour}}$
May 7th 2006, 09:55 AM
Failing Math

question about the answer to the first question.

"Since each one travels to each other at 20 miles per hour the amount of time required is 5 hours."

So, did you add the mph of the two bikes (10mph) together?

Then that would mean that for 10mph and 20mph, I'd start with 30, right?

BTW - Thanks for such fast responses! You're AWESOME!!
May 7th 2006, 10:03 AM
ThePerfectHacker

Quote:

Originally Posted by Failing Math

"Since each one travels to each other at 20 miles per hour the amount of time required is 5 hours."

So, did you add the mph of the two bikes (10mph) together?

Then that would mean that for 10mph and 20mph, I'd start with 30, right?

Correct.

Quote:

Originally Posted by Failing Math

BTW - Thanks for such fast responses! You're AWESOME!!

I know, you can bow down to me if you want to.
May 7th 2006, 10:04 AM
Failing Math

Thanks!

LoL THANKS A BUNCH! I really appreciate it!
May 7th 2006, 01:13 PM
topsquark

Quote:

Originally Posted by ThePerfectHacker

I know, you can bow down to me if you want to.

I'm not sure he was joking about that... ;)

-Dan

(That reminds me of the day I had to TA a class about an hour after my Thesis defense. I didn't tell them specifically how I did. I simply told them that they were now required to call me "Master." :) )