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July 17th, 2014, 02:58 AM   #1
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Find the normal average rate at which Isatu walks to school.

Isatu walks a distance of 1.5km to school everyday; her rate of walking being always constant. On a certain day, owing to ill health, she had to reduce her walking rate by 0.5km/ h and as a result, she took extra 6 minutes to reach the school than usual. Find the normal average rate at which Isatu walks to school.

Time t = 6/60=1/10h
Let her normal walking rate be x.
If she reduce it by 0.5km/h, the walking rate becomes (x-0.5)km/h when she was sick.
I can't go further here, I need help please.
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July 17th, 2014, 05:27 AM   #2
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Velocity is v, time is t. vt=1.5 and (v-0.5)(t+0.1)=1.5
6 min. is 0.1 hours.
We get t=(v-0.5)/5 and we know t=1.5/v.

Rest is easy.

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July 17th, 2014, 08:21 AM   #3
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You forgot change your profile picture.


$\displaystyle \color{blue}{Read \ "Decalogue"...
\\\;\\


and\ let \ you \ read .}$
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July 17th, 2014, 11:20 AM   #4
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Quote:
Originally Posted by aurel5 View Post
You forgot change your profile picture.


$\displaystyle \color{blue}{Read \ "Decalogue"...
\\\;\\


and\ let \ you \ read .}$
Quote:
Originally Posted by tahirimanov View Post
Velocity is v, time is t. vt=1.5 and (v-0.5)(t+0.1)=1.5
6 min. is 0.1 hours.
We get t=(v-0.5)/5 and we know t=1.5/v.

Rest is easy.

Change your profile picture.
Why do you want me to change my profile picture?
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July 17th, 2014, 09:27 PM   #5
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Quote:
Originally Posted by tahirimanov View Post
We get t=(v-0.5)/5 and we know t=1.5/v.
How did you get the two equations above?
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July 18th, 2014, 12:41 AM   #6
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Why do you want me to change my profile picture?


There is no place .



So, it's better let us, profane humanity and unconfessed,

to look calm and safe the sanctity of mathematics.
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July 18th, 2014, 06:23 AM   #7
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(v-0.5)(t+0.1)=1.5 therefor vt+0.1v-0.5t-0.05=1.5 and we know vt=1.5 therefore 0.1v-o.5t-0.05=0 therefore v-5t-0.5=0 and t=1.5/v therefore v-7.5/v-0.5=0

v^2 - 0.5v - 7.5 = 0
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July 19th, 2014, 09:00 PM   #8
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$(v-0.5)(t+0.1)=1.5$ therefor $vt+0.1v-0.5t-0.05=1.5$ and we know$\displaystyle { vt=1.5\ therefore\ 0.1v-0.5t-0.05=0\ therefore\ v-5t-0.5=0 and t=\frac{1.5}{v} \ therefore \ v-\frac{7.5}{v}-0.5=0}$

$v^2-0.5v-7.5= 0$
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$v^2-0.5v-7.5= 0$
I can see clearly that the equation above cannot be solved using factorization method. Or could it be solved using factorization method?
Using quadratic formula, we will come to a point where we have $\displaystyle \frac{0.5\pm\sqrt{30.25}}{2}$
= $\displaystyle \frac{0.5pm5.5}{2}\rightarrow \\
v=3 \;or -2.5$
So normal rate = $3km/h$

Last edited by Chikis; July 19th, 2014 at 09:19 PM.
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July 19th, 2014, 09:15 PM   #9
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-2.5 is also valid solution, if Isatu walks backwards.
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July 19th, 2014, 10:22 PM   #10
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Quote:
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I can see clearly that the equation above cannot be solved using factorization method. Or could it be solved using factorization method?
Yes; multiply by 2:
2v^2 - v - 15 = 0
Now factor:
(2v + 5)(v - 3) = 0
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