Sloan the sloth crawls at 100 meters per hour when crawling uphill, 120 meters per hour when crawling across flat ground, and 150 meters per hour when crawling downhill. One day, Sloan crawls across the forest floor from a tree stump to a spring, and then takes the same route in reverse to return to the tree stump. What was Sloan's average speed during the entire round trip?
I already made an equation:
$average\:speed = \frac{(a+b+c)}{\frac{a}{100}+ \frac{b}{120} + \frac{c}{150} }$
a=total roundtrip distance uphill
b=total roundtrip distance flat ground
c=total roundtrip distance downhill
But idk if this is right or if I solved it right
+118687
On the way there
let the initial uphill bit be a metres
let the flat bit be b metres
and let the downhile bit be c metres
\(speed=\frac{distance}{time}\\ so\\ time=\frac{distance}{speed}\\\)
\(time = \frac{a}{100}+\frac{b}{120}+\frac{c}{150}\quad+ \quad \frac{c}{100}+\frac{b}{120}+\frac{a}{150}\\ time =\frac{1}{10}[ \frac{a+c}{10}+\frac{2b}{12}+\frac{c+a}{15}]\quad\\ time =\frac{1}{10}[ \frac{30(a+c)}{300}+\frac{50b}{300}+\frac{20(c+a)}{300}]\quad\\ time =\frac{1}{3000}[ 30a+30c+50b+20c+20a]\quad\\ time =\frac{1}{3000}[ 50a+50b+50c]\quad\\ time =\frac{1}{60}[ a+b+c]\quad\\\)
\(averaage \;Speed=\frac{distance}{time}\\ averaage \;Speed=\frac{2(a+b+c)}{\frac{1}{60}(a+b+c)}\\ averaage \;Speed=120 \;m/hour \)
LaTex:
time = \frac{a}{100}+\frac{b}{120}+\frac{c}{150}\quad+ \quad \frac{c}{100}+\frac{b}{120}+\frac{a}{150}\\
time =\frac{1}{10}[ \frac{a+c}{10}+\frac{2b}{12}+\frac{c+a}{15}]\quad\\
time =\frac{1}{10}[ \frac{30(a+c)}{300}+\frac{50b}{300}+\frac{20(c+a)}{300}]\quad\\
time =\frac{1}{3000}[ 30a+30c+50b+20c+20a]\quad\\
time =\frac{1}{3000}[ 50a+50b+50c]\quad\\
time =\frac{1}{60}[ a+b+c]\quad\\
+15000 What was Sloan's average speed during the entire round trip?
Hello Guest!
The average speed on sloped terrain is:
\(v_1=\frac{100+150}{2}\ m/h=125\ m/h\)
The average speed on level ground is:
\(v_2=120\ m/h\)
The section of sloping terrain is \(s_1\),
the section of flat terrain is \(s_2\).
The sloan covers the distance in time t.
\(t=\frac{s_1}{v_1}+\frac{s_2}{v_2}=\frac{s_1v_2+s_2v_1}{v_1v_2}\)
\(s=s_1+s_2\)
\(v=\frac{s}{t}=\frac{(s_1+s_2)v_1v_2}{s_1v_2+s_2v_1}\)
The average speed depends on the ratio of the sections.
!
+118687
On the way there
let the initial uphill bit be a metres
let the flat bit be b metres
and let the downhile bit be c metres
\(speed=\frac{distance}{time}\\ so\\ time=\frac{distance}{speed}\\\)
\(time = \frac{a}{100}+\frac{b}{120}+\frac{c}{150}\quad+ \quad \frac{c}{100}+\frac{b}{120}+\frac{a}{150}\\ time =\frac{1}{10}[ \frac{a+c}{10}+\frac{2b}{12}+\frac{c+a}{15}]\quad\\ time =\frac{1}{10}[ \frac{30(a+c)}{300}+\frac{50b}{300}+\frac{20(c+a)}{300}]\quad\\ time =\frac{1}{3000}[ 30a+30c+50b+20c+20a]\quad\\ time =\frac{1}{3000}[ 50a+50b+50c]\quad\\ time =\frac{1}{60}[ a+b+c]\quad\\\)
\(averaage \;Speed=\frac{distance}{time}\\ averaage \;Speed=\frac{2(a+b+c)}{\frac{1}{60}(a+b+c)}\\ averaage \;Speed=120 \;m/hour \)
LaTex:
time = \frac{a}{100}+\frac{b}{120}+\frac{c}{150}\quad+ \quad \frac{c}{100}+\frac{b}{120}+\frac{a}{150}\\
time =\frac{1}{10}[ \frac{a+c}{10}+\frac{2b}{12}+\frac{c+a}{15}]\quad\\
time =\frac{1}{10}[ \frac{30(a+c)}{300}+\frac{50b}{300}+\frac{20(c+a)}{300}]\quad\\
time =\frac{1}{3000}[ 30a+30c+50b+20c+20a]\quad\\
time =\frac{1}{3000}[ 50a+50b+50c]\quad\\
time =\frac{1}{60}[ a+b+c]\quad\\
+37153 Basically the same as Melody's answer:
Time on uphill one way = time downhill coming back
'a' distance uphill is 'a' distance downhill on the way back
c downhill is c uphill on the way back
b is flat distance = 2 b for the roundtrip
total distance = 2 (a+b+c)
(a+c) / 100 + 2 b / (120) + (a+c)/ 150 = time for round trip common denominator 600
[ (6)(a+c) + 10b + 4 (a+c) ] / 600 = time <==simplify
time = 1/60 (a+b+c) remember distance/time = rate
2 (a+b+c) / (1/60(a+b+c) = 120 m/s
