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Sarah walked 10 km into the country. She returned walking 3 km/h slower. The total time for the
trip was 7 hours. How fast did she walk going out to the country?

 May 29, 2020
 #1
avatar+30322 
+2

Let v be speed in km/hr on outward journey.

 

Outward time:  10/v hr

Return time: 10/(v - 3) hr

 

Total time:  10/v + 10/(v - 3) = 7

 

Can you take it from here?
 

 May 29, 2020
edited by Alan  May 29, 2020
 #2
avatar+47 
+4

\(x = {v*t}\)

Vi = initial velocity

T= second time

T= first time

\(10= {Vi* Ti}\)

\(10 = {(Vi-{3km\over h}(tf))}\)

\(Vi*Ti= {(Vi- { 3km\over h}})*tf\)

\(Tf = {7h-Ti}\)

\(Vi*Ti = (Vi-{3km\over h})*(7h-Ti)\)

\(Vi*Ti = ({Vi*7h-ViTi-{3km\over h}}+{3km \over h}*Ti)\)

\(Vi*Ti= {7h*Vi - 21km + {3km\over h}}*Ti\)

\(20km = {7h*Vi - 21km +{ 3km\over h}}*Ti\)

\(41km ={ {70km*h+ { 3km\over h}}*Ti^2\over Ti}\)

\(41km ={ {7h*10km \over Ti}+{3km*Ti\over h}\over Ti}\)

\(0={ {3km*Ti^2\over h} - 41km*Ti+70 km*h }\)

\(0 = {3x^2 -41 x + 70 }\)

This equation gives us two answers:

2, and 35/3

2 = the number of hours in the initial trip

 

 

How fast did she walk going out to the country?

\(5km\over h\)

.
 May 29, 2020

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