# bike race

• Oct 1st 2008, 05:23 PM
yoleven
bike race
Two riders are racing. When A has 6 km to go he is told that B is 4 minutes ahead of him and travelling a constant speed. A speeds up to a rate 2 km/hr faster than B to try to catch him. When A finishes he learns that he lost by 2 minutes.
What was A speed over the last 6 km? Both riders had a constant speed.

I would love some help on this one. I can't get started.
• Oct 2nd 2008, 05:49 AM
ticbol
Let r = rate of B, in km/hr.
So, (r+2) = rate of A in the last 6 km.

Time, t, for A to negotiate that last 6 km,
t = distance/rate = 6/(r+2) hrs

Time, t', for B to negotiate that last 6 km,
t' = 6/r hrs

B is ahead by 4 minutes when A was 6 km from finish. Then A lost by 2 minutes in the race.
That means
t = t' -4min +2min
So,
6/(r+2) = 6/r -2min
6/(r+2) = 6/r -1/30
6/(r+2) = (180 -r)/(30r)
Cross multiply,
6*30r = (r+2)(180 -r)
180r = 180r -r^2 +360 -2r
0 = -r^2 -2r +360
r^2 +2r -360 = 0
r = {-2 +,-sqrt[2^2 -4(1)(-360)]} / 2(1)
r = 18 km/hr or -20 km/hr.

Hence, the speed of A in the last 6 km was 18 +2 = 20 km/hr ----answer.
• Oct 3rd 2008, 03:15 PM
yoleven
My teacher said that was incorrect because in the equation
\$\displaystyle 6/r+2=6/r-1/30\$
the \$\displaystyle 6/r\$ doesn't reflect B in the last \$\displaystyle 6 km\$
She gave the equations

\$\displaystyle 1. tA= tB+2\$

\$\displaystyle 2. vA=vB+2km\$

\$\displaystyle 3. vAtA=6km\$

\$\displaystyle 4. vBtB=6km-4/60\$

She said by manipulating these I could find out A's speed, etc.
Does anyone have any ideas?
• Oct 3rd 2008, 10:27 PM
ticbol
Quote:

Originally Posted by yoleven
My teacher said that was incorrect because in the equation
\$\displaystyle 6/r+2=6/r-1/30\$
the \$\displaystyle 6/r\$ doesn't reflect B in the last \$\displaystyle 6 km\$
She gave the equations

\$\displaystyle 1. tA= tB+2\$

\$\displaystyle 2. vA=vB+2km\$

\$\displaystyle 3. vAtA=6km\$

\$\displaystyle 4. vBtB=6km-4/60\$

She said by manipulating these I could find out A's speed, etc.

With the 4 equations given to you by your teacher, you cannot solve your original question.

Impossible.

Not only that his unts are not consistent, but also her relastionships or "equations" in the first and fourth equations are wrong.

This Math. There should be only one correct answer.
Let us compare it with mine.

------------------------------------
Let us check mine.

I said that vA = 20 Km/hr

That's because I've found out that vB = 18 km/hr.

So, tA = 6 / vA = 6 /20 = 0.3r = 18 minutes.
Meaning, A reaches the finish line in 18 minutes from where he increased his speed.

B is ahead by 4 minutes when A is 6 km from finish.
d = rate*time = 18*(4/60) = 1.2 km.
Meaning, when A is 6 km from finish line, B is only 6 -1.2 = 4.8 km from the finish line.
So, B has to run only
t = d/v = 4.8 /18 = 0.2666667 hrs to reach the finish line.
That is (2.666667 hr)(60min /1hr) = 16 minutes.

Thus, since A takes 18 minutes to finish line while B takes only 16 minutes, then it is clear that A loses by 2 minutes. And that's what your question says.
• Oct 4th 2008, 12:07 AM
CaptainBlack
Quote:

Originally Posted by yoleven
My teacher said that was incorrect because in the equation
\$\displaystyle 6/r+2=6/r-1/30\$
the \$\displaystyle 6/r\$ doesn't reflect B in the last \$\displaystyle 6 km\$
She gave the equations

\$\displaystyle 1. tA= tB+2\$

\$\displaystyle 2. vA=vB+2km\$

\$\displaystyle 3. vAtA=6km\$

\$\displaystyle 4. vBtB=6km-4/60\$

She said by manipulating these I could find out A's speed, etc.
Does anyone have any ideas?

ticbols approach is right (I've not checked his algebra and arithmetic, but it is usually reliable but you should check it anyway), he is also right about the inconsistency of the units you say your teacher gave, in equation 1. all time units should be in hours .

Equation 4. cannot be right you cannot subtract time from distance to get a distance.

Now it is possible that you teacher had in mind a method that looks different from ticbols, but if it is right it will be equivalent, there is usually more than one way of tackling a problem.

You should look at ticbols solution and try to understand why he does what he does, then you can explain to your teacher what is going on. The solution is posted for you to learn how to deal with problems like this, not just for you to hand in as your own work.

RonL
• Oct 5th 2008, 12:21 PM
yoleven
I didn't try to pass off his work as mine. The teacher discussed the question in class and it was totally way off from what I understood from ticbols.
I like his approach much better.
When I approached her after, she said flat out that it was wrong. She didn't give the answer and at this point, I'm confused and I don't really know what she was getting at.
I have been working on this off and on and I am so sick of this question.
I went to her office and when I asked for advice on this and other questions like these, she just said to practice.
• Oct 6th 2008, 01:29 AM
ticbol
Quote:

Originally Posted by yoleven
I didn't try to pass off his work as mine. The teacher discussed the question in class and it was totally way off from what I understood from ticbols.
I like his approach much better.
When I approached her after, she said flat out that it was wrong. She didn't give the answer and at this point, I'm confused and I don't really know what she was getting at.
I have been working on this off and on and I am so sick of this question.
I went to her office and when I asked for advice on this and other questions like these, she just said to practice.

Umm, your teacher is , I'd say, a typical Math teacher or Math "expert". She cannot accept mistakes. :-)
They think they are the best.

It is easy to tell students the way to solve a problem...what with all those impressive-sounding formulas and Math know-how. But for me, if I don't see the answer, I just, ah, meh at that way of teaching.
Show me the eventual answer, even without showing how you arrived at it, after all those teaching methods....because there should be only one correct answer (or sets of answers)....then I will appreciate or try to check if that answer is correct or not. Or, at least, if that answer will be the same as mine. (Mine could be wrong.) Then I will believe you. No answer, I will leave you alone.

I had some Math teachers in High School and in college too that were like that. They'd rather send me out of the classroom when I insisted my answers were correct, with me showing or trying to show my solutions in details. :-)

In your case with your teacher now, do not believe her if you think she is wrong. If you don't want to be told to get out of her classroom, then just shut your mouth or don't cross her.
Next school year, or next semester, you'd have another teacher.