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?
\(x = {v*t}\)
Vi = initial velocity
Tf = second time
Ti = 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 = the number of hours in the initial trip
How fast did she walk going out to the country?
\(5km\over h\)