Derive equation of runner

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September 19th 2010, 11:10 AM
webguy

Derive equation of runner

Hi,

An athlete usually runs 80km at a steady speed of $v$ 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 $\frac{80}{v}+\frac{8}{3}=\frac{160}{2v-5}$

Here's what I did which unfortunately is wrong:

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

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

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

Quote:

Originally Posted by webguy

Hi,

An athlete usually runs 80km at a steady speed of $v$ 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 $\frac{80}{v}+\frac{8}{3}=\frac{160}{2v-5}$

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

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

clear the decimal in the last fraction ...

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

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

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

numerators form the equation ...

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

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

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

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

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

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
September 19th 2010, 12:40 PM
webguy

Quote:

Originally Posted by skeeter

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

clear the decimal in the last fraction ...

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

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

Also, why is my answer so incorrect. The logic of my first step seems correct in my mind; is it not?
September 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 $\frac{2}{2}$ .

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.

...
September 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?
September 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 ...

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

$\displaystyle \frac{80 \, km}{v \, km/hr} \cdot \frac{60 \, min}{hr}$ = that same time above in minutes
September 19th 2010, 03:10 PM
webguy

So it should be $\frac{4800}{60v}$ ?
September 19th 2010, 03:47 PM
skeeter

Quote:

Originally Posted by webguy

So it should be $\frac{4800}{60v}$ ?

no
September 20th 2010, 02:39 PM
webguy

Quote:

Originally Posted by skeeter

no

Could you elaborate?
September 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.
September 21st 2010, 01:16 PM
webguy

Quote:

Originally Posted by skeeter

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

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

So my original should be:

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

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.