Calculating Average Speed of a Car Round Trip

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Homework Help Overview

The discussion revolves around calculating the average speed of a car during a round trip, where the car travels uphill at a speed of 32 km/h and downhill at 66 km/h. The original poster expresses confusion due to the absence of time and distance information in the problem statement.

Discussion Character

  • Exploratory, Assumption checking

Approaches and Questions Raised

  • Participants explore the implications of missing time information and question whether the problem is solvable as presented. Some suggest considering specific distances to derive a pattern for average speed calculations.

Discussion Status

Participants are actively engaging with the problem, with some providing guidance on how to approach the average speed calculation by suggesting the use of specific distances. There is a recognition of the importance of the word "returns" in the context of the problem.

Contextual Notes

There is an ongoing discussion about the assumptions made regarding distance and time, with some participants suggesting that the problem may lack necessary information while others argue that it can be approached with hypothetical distances.

tigerlili
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Homework Statement


A car travels up a hill at a constant speed of 32 km/h and returns down the hill at a constant speed of 66 km/h. Calculate the average speed (in km/h) for the round trip.

Homework Equations



average speed= total distance/ delta time is the only equation i know for it

The Attempt at a Solution



there is no time, or actual distance given in this question, so i got confused and just tried to find the average from adding up the two speeds.. obviously that didn't work- this seems like a really simple problem, so where am i going wrong?
 
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I don't think your doing anything wrong. I think the problem is missing the time information.

Thanks
Matt
 
but... it's online homework. that was just the given problem :/
 
Well, I have solved many dynamics problems and this problem is missing the information needed to solve it. Can you contact the teacher/instructor who posted it?

Thanks
Matt
 
CFDFEAGURU is, I fear, leading you astray. There is no missing fact, although there is a very important word, "returns". That means the distance up the hill is the same as the distance down the hill.
 
Whoops LOL. I missed that word.

Sorry to the OP.

Thanks
Matt
 
But.. it's average speed, not velocity.. so, direction doesn't matter, i thought? and you can't just average the speeds they give you.. and there's no time given :/
 
Suppose the distance is 1km. Work out average speed. 2km? 5km? See a pattern?

Now try x km.
 
so.. like, suppose 1 km is the distance

so do 32 km/hr / 1 km = 32 hr^-1 ,etc?

then what? :/ I'm sorry, i just really need this to be actually taught to me..
 
  • #10
Well, we know that
s=\frac{d}{t}
s=\frac{d_1+d_2}{t_1+t_2}
d1 and t1 are for the first have of the trip @ 32 km/h and d2 and t2 are for the other half at 66km/h.
Now you need to find what each of those are equal to.
 
  • #11
but, how..
 
  • #12
d_1=d_2=x
v_1 = \frac{x}{t_1}
v_2 = x/t_2
Solve for the t's and plug into the speed equation in my last post. Then do some algebra.
 
  • #13
so i have s= 2x/(x/32 + x/66)

but.. if we don't know what s is and we don't know what x is
what comes next?
 
  • #14
s=\frac{2x}{\frac{x}{32}+\frac{x}{66}}
If you find a common denominator for the lowed half of the equation the x's will cancel out.
 
  • #15
yes, of course you're right

thank you very much for your help, i really appreciate it
 

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