SOLUTION: "Two cyclists travel on a straight path towards each other each travelling at a constant speed of 10mph. The cyclists start of 10 miles apart. If a fly starts at the front of one c

Algebra ->  Customizable Word Problem Solvers  -> Travel -> SOLUTION: "Two cyclists travel on a straight path towards each other each travelling at a constant speed of 10mph. The cyclists start of 10 miles apart. If a fly starts at the front of one c      Log On

Ad: Over 600 Algebra Word Problems at edhelper.com


   



Question 733415: "Two cyclists travel on a straight path towards each other each travelling at a constant speed of 10mph. The cyclists start of 10 miles apart. If a fly starts at the front of one cyclists and keeps flying backwards and forward until the two cyclists crash, how far will the fly fly give that the fly travels at 1mph?"
I'm not sure where to start... please could you give a brief explanation of how to work it out if possible.
Thanks if you can help

Found 3 solutions by lynnlo, ikleyn, MathTherapy:
Answer by lynnlo(4176) About Me  (Show Source):
You can put this solution on YOUR website!
D=DISTANCE,,,,,R=RATE,,,,,,T=TIME
CYCLISTS A========10MPH
CYCLISTS B========10MPH
MULTIPLY THEIR TRAVEL SPEED TIMES THE DISTANCE

Answer by ikleyn(53427) About Me  (Show Source):
You can put this solution on YOUR website!
.
"Two cyclists travel on a straight path towards each other each travelling at a constant speed of 10mph.
The cyclists start of 10 miles apart. If a fly starts at the front of one cyclists and keeps flying backwards
and forward until the two cyclists crash, how far will the fly give that the fly travels at 1 mph?"
I'm not sure where to start... please could you give a brief explanation of how to work it out if possible.
Thanks if you can help
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~


Without calculations, it is clear that the cyclists will meet each other in half-an-hour, or 30 minutes.


All this time the fly will move at the speed of 1 mile per hour.


So, the total travel distance for the fly is 0.5 of an hour times 1 mile per hour, or

    0.5 * 1 = 0.5  of a mile.


Thus the problem is solved completely.


ANSWER.  The traveled distance for the fly is  0.5 of a mile.

Solved.

------------------------

The post-solution note - - - read it attentively


As the problem is posed in your post, the fly will not move backward - it will move forward ONLY
during this half-an-hour before the cyclists will meet each other.

It is because in your problem the speed of the fly, which is given as 1 mph, is tooooo small
in order for the fly could reach the other cyclist before the first cyclist will reach the second one.

So, you either incorrectly formulate the problem or copied it from wrong source.

But in any case, the real movement of the fly in this problem is not as you describe in your post.

In other terms, different parts of your post are not consistent.

I am not surprised, because at this forum I observe such phenomena every day and even several times per day.


Visitors write something or compose something or copy-paste something, but quite rarely
they do understand what they write or what they compose or what they copy-paste.



Answer by MathTherapy(10587) About Me  (Show Source):
You can put this solution on YOUR website!
"Two cyclists travel on a straight path towards each other each travelling at a constant speed of 10mph.
The cyclists start of 10 miles apart. If a fly starts at the front of one cyclists and keeps flying
backwards and forward until the two cyclists crash, how far will the fly fly give that the fly travels at 1mph?"

I'm not sure where to start... please could you give a brief explanation of how to work it out if possible.
Thanks if you can help

The problem should NOT READ, "flying backwards and forward", BUT  "flying BACK and FORTH", instead.  

Distance cyclists are apart: 10 miles
Speed each cyclist travels: 10 mph

Let time taken for cyclists to meet, be T
Then we get the following DISTANCE equation: 10T + 10T = 10
                                                   20T = 10
                 Time taken for cyclists to meet, or T = 10%2F20+=+1%2F2hr
Time fly spent flying is ALSO 1%2F2hr

Since fly spent 1%2F2hr flying, at 1 mph, distance fly flew = .