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

AI Thread Summary
The discussion revolves around a physics homework problem involving average speed and average velocity during a trip. The user initially misinterprets the questions, leading to confusion between average speed and average velocity, with calculations yielding incorrect results. Key points include the correct formulas for average speed and average velocity, emphasizing the importance of understanding concepts rather than just applying equations. The user also seeks recommendations for introductory physics textbooks, indicating a desire for further learning. Overall, the thread highlights common pitfalls in physics problem-solving and the need for clarity in textbook questions.
Const@ntine
Messages
285
Reaction score
18

Homework Statement


John is entering Steward, and drives West, with a Uav of 88 km/h.
He arrives at Aurora, and realizes 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!
 
Physics news on Phys.org
Question is written incorrectly. Part a asks for average speed. Part b means to ask for magnitude of average velocity
 
PhanthomJay said:
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!
 
Darthkostis said:
I get ~21 km/h^2, whereas he has "32 km/h"
I get 32. Please post your working.
 
haruspex said:
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.
 
Darthkostis said:
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,
 
haruspex said:
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.

haruspex said:
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

haruspex said:
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 realize 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!
 
Darthkostis said:
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.
 
haruspex said:
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.
 

Similar threads

Back
Top