+0

# HELP! SO CONFUSED!!!

0
148
3

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?

$average\:speed = \frac{(a+b+c)}{\frac{a}{100}+ \frac{b}{120} + \frac{c}{150} }$

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

Mar 4, 2022
edited by Guest  Mar 4, 2022

#2
+2

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:

Mar 4, 2022

#1
+1

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. !

Mar 4, 2022
edited by asinus  Mar 4, 2022
#2
+2

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:

Melody Mar 4, 2022
#3
+1

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

Mar 4, 2022