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zello
Group Title
Nicole hiked up a mountain trail and camped overnight at the top. The next day she returned down the same trail. Her average rate traveling uphill was 2.6 kilometers per hour and her average rate downhill was 3.9 kilometers per hour. If she spent a total of 12 hours hiking, how long was the trail?
 2 years ago
 2 years ago
zello Group Title
Nicole hiked up a mountain trail and camped overnight at the top. The next day she returned down the same trail. Her average rate traveling uphill was 2.6 kilometers per hour and her average rate downhill was 3.9 kilometers per hour. If she spent a total of 12 hours hiking, how long was the trail?
 2 years ago
 2 years ago

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geoffb Group TitleBest ResponseYou've already chosen the best response.1
You know the trail is the same length going up as it is going down. Therefore, her total distance travelled is \(2d\). \[ v = \frac{d}{t}\] \[t_{\text{up}} = 12  t_{\text{down}}\] This should get you started.
 2 years ago

zello Group TitleBest ResponseYou've already chosen the best response.0
dw:1354392763687:dw
 2 years ago

zello Group TitleBest ResponseYou've already chosen the best response.0
I didn't get it =(
 2 years ago

geoffb Group TitleBest ResponseYou've already chosen the best response.1
Yeah, that looks right. Now you just need to make it into an equation.
 2 years ago

geoffb Group TitleBest ResponseYou've already chosen the best response.1
Well, if \(d_{total} = (vt)_{total}\), and \(d_{total} = d_{up} + d_{down}\), then \(d_{total} = (vt)_{up} + (vt)_{down}\)
 2 years ago

zello Group TitleBest ResponseYou've already chosen the best response.0
So, the answer is 36 ?
 2 years ago

geoffb Group TitleBest ResponseYou've already chosen the best response.1
Said another way: \[d_{up} = (vt)_{up}\] \[d_{down} = (vt)_{down}\] \[d_{up} + d_{down} = 2d = (vt)_{up} + (vt)_{down}\]
 2 years ago

geoffb Group TitleBest ResponseYou've already chosen the best response.1
I haven't calculated it, so I'm not sure. One sec.
 2 years ago

zello Group TitleBest ResponseYou've already chosen the best response.0
2.6 X +3.9(12X)= the answer ?
 2 years ago

geoffb Group TitleBest ResponseYou've already chosen the best response.1
No, the answer is not 36.
 2 years ago

geoffb Group TitleBest ResponseYou've already chosen the best response.1
Okay, let's start over. First, you know that the distance going up is the same as it is going down. Therefore, you can make \(d_{up}\) = \(d_{down}\).
 2 years ago

geoffb Group TitleBest ResponseYou've already chosen the best response.1
Let's call time \(t\) instead of \(x\).
 2 years ago

geoffb Group TitleBest ResponseYou've already chosen the best response.1
So, since \(d_{up} = d_{down}\), \((vt)_{up} = (vt)_{down}\)
 2 years ago

zello Group TitleBest ResponseYou've already chosen the best response.0
what is the (vt) ?
 2 years ago

geoffb Group TitleBest ResponseYou've already chosen the best response.1
\(vt\) is velocity times time.
 2 years ago

geoffb Group TitleBest ResponseYou've already chosen the best response.1
We're just plugging in \(vt\) for \(d\).
 2 years ago

geoffb Group TitleBest ResponseYou've already chosen the best response.1
Let's use the formula \((vt)_{up} = (vt)_{down}\). Can you plug in your values? Let's say that \(t\) represents time going up. Therefore, \(12  t\) would be time going down.
 2 years ago

geoffb Group TitleBest ResponseYou've already chosen the best response.1
Also, remember that \((vt)_{up}\) is the same as saying \(v_{up} \times t_{up}\). Same goes for down.
 2 years ago

zello Group TitleBest ResponseYou've already chosen the best response.0
oh okay ,so you're saying that the Distance is the Rate*Time which is \[d=vt\] ,right?
 2 years ago

geoffb Group TitleBest ResponseYou've already chosen the best response.1
Yes, exactly.
 2 years ago

geoffb Group TitleBest ResponseYou've already chosen the best response.1
We can equate rate*time up to rate*time down, because they are both the same trail (i.e., same distance).
 2 years ago

geoffb Group TitleBest ResponseYou've already chosen the best response.1
That's what makes it possible to solve the equation.
 2 years ago

zello Group TitleBest ResponseYou've already chosen the best response.0
\[2.6x+3.9(12x) \]
 2 years ago

geoffb Group TitleBest ResponseYou've already chosen the best response.1
You're adding them together. You need to make them equal to each other.
 2 years ago

zello Group TitleBest ResponseYou've already chosen the best response.0
I think i got it
 2 years ago

geoffb Group TitleBest ResponseYou've already chosen the best response.1
Because \(2.6x = d = 3.9(12x)\)
 2 years ago

geoffb Group TitleBest ResponseYou've already chosen the best response.1
Solve for \(x\) (time up the hill), then plug it into your original formula to solve for \(d\).
 2 years ago

zello Group TitleBest ResponseYou've already chosen the best response.0
\[2.6x=3.9(12x) \] \[2.6x=46.83.9x \] \[46.8=6.5x \] \[x=7.2\]
 2 years ago

geoffb Group TitleBest ResponseYou've already chosen the best response.1
Good! So what does \(x\) represent?
 2 years ago

zello Group TitleBest ResponseYou've already chosen the best response.0
time up the hill ?
 2 years ago

geoffb Group TitleBest ResponseYou've already chosen the best response.1
Yes, exactly!
 2 years ago

geoffb Group TitleBest ResponseYou've already chosen the best response.1
So now you can plug it into your distance formula (remember, \(d = vt\)) and solve for \(d\).
 2 years ago

geoffb Group TitleBest ResponseYou've already chosen the best response.1
You also know that time down the hill is \(12  x\), which you can calculate now to 4.8. You can plug *that* into your downhill formula, and you should get the same answer for \(d\). That will help prove your answer is right.
 2 years ago

zello Group TitleBest ResponseYou've already chosen the best response.0
\[d=vt\]\[3.9*4.8=18.72\] \[d=18.72\]
 2 years ago

geoffb Group TitleBest ResponseYou've already chosen the best response.1
Excellent. And, conversely, \(2.6 \times 7.2 = 18.72 \text{ km}\), so you know your answer is right.
 2 years ago

zello Group TitleBest ResponseYou've already chosen the best response.0
Thank you so much !
 2 years ago

geoffb Group TitleBest ResponseYou've already chosen the best response.1
No problem. :)
 2 years ago
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