I can't quite get this question... (|Uav|)

  • #1
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Homework Statement


John is entering Steward, and drives West, with a Uav of 88 km/h.
He arrives at Aurora, and realises he took a wrong turn.
He turns around and drives East, towards York, with a Uav' of 79 km/h.
Steward-Aurora is 76 km & and Aurora-York is 34 km.

For the whole trip (S -> A -> Y) find:

a) What's the average metre (meaning the positive value of a quantity, eg | -70| = 70) of his speeds (U)?
b) What's the metre of his average speed?

Homework Equations



Uav = (Xfinal - Xstarting)/(tfinal - tfirst)
X=U*t

The Attempt at a Solution



S->A: [/B]Ssa = |Usa|*tsa <=> 76 km = 88 km/h*tsa <=> tsa = 0,86h
A->Y: Say = |Uay|*tay <=> 34 km = 79 km/h*tay <=> tay = 0,43 h

Sc = (76 + 34) km = 110 km
tc = (0,86 + 0,43) = 1,29 h

  • Okay, so, the answers are:

a) 85 km/h

b) 32 km/h

Personally, I think he got them mixed up. The metre of his average speed (b) if I'm not mistaken, should be: |Uav| = Sc/tc = 110 km/ 1,29 h ~ 85 km/h

So, it's just a typo, right? Well, thing is, for the life of me, I can't understand what question (a) means. I translated it here as well as I could. So, any ideas?

Any kind of help would be appreciated!
 

Answers and Replies

  • #2
PhanthomJay
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Question is written incorrectly. Part a asks for average speed. Part b means to ask for magnitude of average velocity
 
  • #3
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Question is written incorrectly. Part a asks for average speed. Part b means to ask for magnitude of average velocity
Well, that does make sense. I get ~21 km/h^2, whereas he has "32 km/h", but I suppose that'd be another typo. Still, such problems really shouldn't be present in textbooks. I mean, wrong answers I get, but wrong questions as well?

Either way, thanks a ton for the help!
 
  • #5
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I get 32. Please post your working.
Well, I just went with the definition that's in the book:

aav = (U2 -U1)/(t2 - t1)

But since he's asking for the meter/positive value, I went with:

|aav| = | (U2 - U1) / (t2 - t1) |

Problem is, it was early in the morning, and I put the wrong values in there (U2 = 88 km/h , U1 = 79 km/h, t1= 0,86 s & t1 = 0,43 s). So yeah, my result was wrong.

Now, he's asking to find that for the whole trip, so I figured I'd go with U2 = 79 km/h (his Uav when he's returning), t2 = 1,29 h (the whole duration), t1 = 0 h & U1 = 88 km/h (his Uan in the beginning), but that's still wrong.

So, yeah, I'm kinda stumped on this one (and I'm at Uni now (on a break) so I don't have the time to currently investigate more). I'd be grateful if I could see your solution, as I'm obviously missing something.
 
  • #6
haruspex
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Well, I just went with the definition that's in the book:
aav = (U2 -U1)/(t2 - t1)
Such are the perils of using equations without understanding what they mean.
That equation gives the average acceleration when a body goes from velocity U1 at time t1 to velocity U2 at time t2. There is nothing in this question about accelerations.

The equations you need are
Average speed = total length of path / total time taken
Average velocity = net displacement (a vector) / total time taken,
 
  • #7
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Such are the perils of using equations without understanding what they mean.
That equation gives the average acceleration when a body goes from velocity U1 at time t1 to velocity U2 at time t2. There is nothing in this question about accelerations.
Well, in my first post my main question was that I didn't get what exactly the questions themselves were asking, so there's that. After that, yeah, I misread PhanthomJay's answer, and instead of velocity I took it as accelaration (twas about 7 in the morning here). I get what you mean though.

The equations you need are
Average speed = total length of path / total time taken
Okay, so the usual Uav = (76+34) km / 1,29 h = 85 km/h

Average velocity = net displacement (a vector) / total time taken,
First off I don't know what net displacement means, but after searching it a bit on the internet, instead of adding the | | of each distance, I view them as vectors, meaning that I'll have to add them while taking into account which direction I've set as the positive one.

So, in my case, let's say that East to West is the positive, and West to East is the negative direction. Therefore, Seward-Aurora becomes +76 km, and Aurora-York becomes -34 km. Thus:

U = (+76 - 34) km/ 1,29 h = 32.5 km/h

I'm working with a translated book, so certain things I either don't know, or don't realise what they actually are, due to the use of different symbols and definitions. Either way, thanks a ton for the help, I really appreciate it!

PS: Are there any other good textbooks that anyone could recommend for 1st Semester Physics (torque, vocation, speed, relativity, balance, thermodynamics, etc, etc). It'd be of great help!
 
  • #8
haruspex
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PS: Are there any other good textbooks that anyone could recommend for 1st Semester Physics (torque, vocation, speed, relativity, balance, thermodynamics, etc, etc). It'd be of great help
I cannot help with that, and you'd probably do better to post this a new thread, not on a homework forum.
 
  • #9
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I cannot help with that, and you'd probably do better to post this a new thread, not on a homework forum.
Fair enough. Thanks for the help anyway.
 

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