# Derive equation of runner

• Sep 19th 2010, 11:10 AM
webguy
Derive equation of runner
Hi,

An athlete usually runs 80km at a steady speed of $\displaystyle v$km per hour. He decides to reduce his speed by 2.5km per hour resulting in his run taking an extra 2 hours and 40 minutes.
Derive $\displaystyle \frac{80}{v}+\frac{8}{3}=\frac{160}{2v-5}$

Here's what I did which unfortunately is wrong:

$\displaystyle \frac{80}{60(v-2.5)}-\frac{80}{60v}=160$ (Converted it to minutes)
$\displaystyle \frac{80}{60(v-2.5)}=160+\frac{80}{60v}$
$\displaystyle \frac{80}{60v-150}=160+\frac{80}{60v}$
$\displaystyle 80=160(60v-150)+\frac{80(60v-150)}{60v}$
$\displaystyle 80=9600v-24000+\frac{80(60v-150)}{60v}$
$\displaystyle 24080-9600v=\frac{80(60v-150)}{60v}$
$\displaystyle 60v(24080-9600v)=80(60v-150)$
$\displaystyle 1444800v-576000v^2=4800v-12000$
$\displaystyle 576000v^2-1440000v-12000=0$
$\displaystyle 48v^2-120v-1=0$

How on Earth did I get it so far from the answer?

Any help is appreciated (Wink)
• Sep 19th 2010, 11:55 AM
skeeter
Quote:

Originally Posted by webguy
Hi,

An athlete usually runs 80km at a steady speed of $\displaystyle v$km per hour. He decides to reduce his speed by 2.5km per hour resulting in his run taking an extra 2 hours and 40 minutes.
Derive $\displaystyle \frac{80}{v}+\frac{8}{3}=\frac{160}{2v-5}$

(time in hrs it normally takes) + (2 hrs + 40 min) = (longer time when he runs slower)

$\displaystyle \frac{80}{v} + \frac{8}{3} = \frac{80}{v-2.5}$

clear the decimal in the last fraction ...

$\displaystyle \frac{80}{v} + \frac{8}{3} = \frac{160}{2v-5}$

common denominator is $\displaystyle 3v(3v-5)$ ...

$\displaystyle \frac{80 \cdot 3(2v-5)}{3v(2v-5)} + \frac{8v(2v-5)}{3v(2v-5)} = \frac{160 \cdot 3v}{3v(2v-5)}$

numerators form the equation ...

$\displaystyle 240(2v-5) + 8v(2v-5) = 480v$

$\displaystyle 30(2v-5) + v(2v-5) = 60v$

$\displaystyle 60v - 150 + 2v^2 - 5v = 60v$

$\displaystyle 2v^2 - 5v - 150 = 0$

$\displaystyle (2v + 15)(v - 10) = 0$

only solution which works in the context of the problem is v = 10 km/hr

at 10 km/hr, takes 8 hrs to run 80 km

at 7.5 km/hr, takes 10 hrs 40 min
• Sep 19th 2010, 12:40 PM
webguy
Quote:

Originally Posted by skeeter
$\displaystyle \frac{80}{v} + \frac{8}{3} = \frac{80}{v-2.5}$

clear the decimal in the last fraction ...

$\displaystyle \frac{80}{v} + \frac{8}{3} = \frac{160}{2v-5}$

How did you do that step? It looks like you only multiplied one side by $\displaystyle \frac{2}{2}$.

Also, why is my answer so incorrect. The logic of my first step seems correct in my mind; is it not?
• Sep 19th 2010, 01:01 PM
skeeter
Quote:

Originally Posted by webguy
How did you do that step? It looks like you only multiplied one side by $\displaystyle \frac{2}{2}$.

correct ... 2/2 = 1 ... multiplication by 1 changes nothing.

Also, why is my answer so incorrect. The logic of my first step seems correct in my mind; is it not?

your "conversion" to minutes (which is not necessary to begin with) is incorrect. note that every term in the first equation represents time in hrs.

conversion to minutes requires multiplying each term by 60 min/hr.

...
• Sep 19th 2010, 02:03 PM
webguy
But I converted every hour to minutes. Even though I was working in minutes I don't see how what I did was wrong. Would you mind pointing out the specific flaws of my answer?
• Sep 19th 2010, 02:22 PM
skeeter
Quote:

Originally Posted by webguy
But I converted every hour to minutes. Even though I was working in minutes I don't see how what I did was wrong. Would you mind pointing out the specific flaws of my answer?

no, you didn't ... I told you in my previous post that each term is a time in hours.

specifically, using the first term of the original equation ...

$\displaystyle \displaystyle \frac{80 \, km}{v \, km/hr}$ = a specific amount of time in hours

$\displaystyle \displaystyle \frac{80 \, km}{v \, km/hr} \cdot \frac{60 \, min}{hr}$ = that same time above in minutes
• Sep 19th 2010, 03:10 PM
webguy
So it should be $\displaystyle \frac{4800}{60v}$?
• Sep 19th 2010, 03:47 PM
skeeter
Quote:

Originally Posted by webguy
So it should be $\displaystyle \frac{4800}{60v}$?

no
• Sep 20th 2010, 02:39 PM
webguy
Quote:

Originally Posted by skeeter
no

Could you elaborate?
• Sep 20th 2010, 03:30 PM
mr fantastic
Quote:

Originally Posted by webguy
Could you elaborate?

Maybe someone with the time and inclination will elaborate in the fullness of time. In the meantime, you need to carefully reflect on the help you've already been given and show that you have taken on board this help.
• Sep 21st 2010, 01:16 PM
webguy
Quote:

Originally Posted by skeeter
$\displaystyle \displaystyle \frac{80 \, km}{v \, km/hr} \cdot \frac{60 \, min}{hr}$ = that same time above in minutes

$\displaystyle 60\times(\frac{80}{v})$ which gives $\displaystyle \frac{4800}{v}$.

So my original should be:

$\displaystyle \frac{4800}{v-2.5}-\frac{4800}{v}=160$

If this is wrong feel free to say with no explanation. I'll end this thread here at that point since I doubt I'll ever understand. Unfortunately I fail to understand why multiplying the hours by 60 minutes and dividing into the distance is incorrect, but multiplying the distance by the minutes is correct.