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 Wilder7bc Jan 18th 2010 06:06 PM

Help with some Basic Word problems Ch 2 book Physics

First off I would like to say hi to everyone!

I am taking Engineering Physics and right now I down right suck at it and simple things seem to confuse me. Its been 13 years since any math class which the last one was Calculus I. So as you can see I am pretty rusty and trying to teach myself math again (and believe me I forget some really simple rules at times) as I move through physics. I have bought some videos from Math Tutor, on Physics and Math to try and help but its tough fight at this early stage.

just for the record we are graded on homework for doing it not on its content. Meaning the professor has the solution guide which is free for us to use in the library (soon as it gets in they changes books this semester).

So when I ask for help on book questions this is not breaking the rules as we are not graded on this portion at least in that manner they are just considered practice problems not if they are right or wrong.

My first question seems so simple but I cannot figure out how the book has the the answer it is as follows:

A person walks first at a constant speed of 5.00m/s along a staright line from a point A to B and then back along the line from B to A at a constant speed of 3.00m/s.

(a) What is her average speed over the entire trip?

(b) what is her Average velocity over the entire trip?

(a) = 3.75m/s (I dont know how they got this answer its killing me!

(b) = 0 this I knew the answer to before looking for the following reason in the book they give the following example:

A runner who runs a distance d of more than 40km and yet ends up at her starting point. her total displacement is zero, so her average velocity is zero!

I guess in my thinking (b) means (xf - xi)/(tf - ti) since its displacement they started out at zero then ended back up at zero and zero decided by anything is zero. Thus velocity is zero.

Where as if it was average speed has no direction and is always expressed as a positive number xf = some amount and xi = 0 So it would be the total amount traveled decided by time. So I think I have question (b) down.. I hope...

However on (a) I am stumped! its irritating becaues it seems like it has to be so simple but I am missing it :(

Thanks,

Brian

 physicsquest Jan 18th 2010 11:32 PM

The average speed is defined as the total distance / total time.

Let distance between A and B = d. Then total distance to and fro = 2d.

The time one way = d/5 and the other way = d/3 . Total time therefore is

d/5 + d/3 = 8d/15.

Avg speed = 2d / (8d/15) = 3.75

 Wilder7bc Jan 19th 2010 04:50 AM

hhmm

Quote:
 Originally Posted by physicsquest (Post 11822) The average speed is defined as the total distance / total time. Let distance between A and B = d. Then total distance to and fro = 2d. The time one way = d/5 and the other way = d/3 . Total time therefore is d/5 + d/3 = 8d/15. Avg speed = 2d / (8d/15) = 3.75
I certainly see how it works with the math....

However I dont see the relationship clearly or the logic behind it.

I am looking at something the wrong way here. What is throwing me off is I know Velocity = 5m/s A to B and then 3m/s from B to A.

you put 5, and 3 in the denominator. Thats what I am having problems seeing. I thought that they should go in the numerator. As they give us the formula for V= d/t we only know the velocity we dont know the time or the distance so we cannot change it around and solve.

So I am trying to look at this from the wrong angle as I see how you did this with math but I am not making the connection to the relationship clearly.

Anyway to help me open my eyes on this I feel a bit loss on it as I would have never made that jump to come up with that answer. To me it seems like the random and the numerator was just put into the denominator and the math happened to work. I guess I cannot see in common sense, layman terms what you did in a way that makes sense and only understand that the math puts out the correct value.

arrrggg so frustrating :(

Brian

 Wilder7bc Jan 19th 2010 05:54 AM

perhaps...

Does this perhaps have something to do with the fact that

Velocity is the derivative of position? which could be seen as distance?

Here is another problem it seems like I am confused as its very similiar:

#5

A position - time graph for a particle moving along the X axis is shown in Figure p2.5

(a). find the average velocity in the time interval 1.50s to t= 4.00s

I am not sure how they got this answer either...

I know that change distance / change in time = velocity.

I know that if you take t(final) - t(initial) it gives you change in time. Which that = 2.5s

I think this is similiar to the problem above and I think I am not making some perception or leap that I need to make to put this together something in the relationship remains hidden to me...

Grrrrrr

 YellowPeril Jan 19th 2010 07:24 AM

Help with some Basic Word problems Ch 2 book Physics

Wilder7bc,

I will try to help you out by stating it another way,

Speed = Distance / Time.

The distance is 2d. We know the velocity for every part of the journey. Therefore we can work out the time. Lets say we take the average of 3 and 5 to give 4m/s, we can see that this is not the correct answer and it will only be correct if the velocity was 4m/s in both cases. To see this say the speed to the target is 7.9m/s and the speed returning is 0.1m/s then the average speed is 2d / (d/7.9+d/0.1) = 0.1975 which is a lot closer to 0.1 than 4. In other words, the slower speed has a greater influence on the average than the faster speed and that is why we use a weighted average rather than just the average.

 Wilder7bc Jan 19th 2010 08:53 AM

why?

Quote:
 Originally Posted by YellowPeril (Post 11833) Wilder7bc, I will try to help you out by stating it another way, Speed = Distance / Time. The distance is 2d. We know the velocity for every part of the journey. Therefore we can work out the time. Lets say we take the average of 3 and 5 to give 4m/s, we can see that this is not the correct answer and it will only be correct if the velocity was 4m/s in both cases. To see this say the speed to the target is 7.9m/s and the speed returning is 0.1m/s then the average speed is 2d / (d/7.9+d/0.1) = 0.1975 which is a lot closer to 0.1 than 4. In other words, the slower speed has a greater influence on the average than the faster speed and that is why we use a weighted average rather than just the average.
I think its this "Working out the time part" I am not getting you say distance is 2d

(I am guessing thats only for this particular equation as they travel so much distance one way then travel the same distance back so your making a common sense approach that 2x unknown variable gives us the full distance traveled.)

Ok then you want to divide that by t= time but we dont have time so since we are dividing time we do the opposite and multiple on the other side of the equation.

Then since since speed actually = velocity and its now being multiplied by time we can take it and divide on the other side and its then that we divide 2d by speed which as it turns out is (3m/s + 5m/s) ? I am guessing we make it addition instead of the normal V(final) - V(initial) because we are wanting the sum?

However its still odd because on the bottom you have d/my velocity numbers of 5 instead of just doing example: 2d/(vf - vi)

I am still confused as I learned these formula for x potion, velocity and then acceleration.

I then know I can solve for the unknown using basic algebra as long as I do not have two unknown variables.

However this stuff seems to be confusing me as its not logical laid out steps to arrive to what I need, or at least I am not seeing it that way yet.

A big part that confuses me is how (d/(d/v)) and more specifically the (d/v) in the denominator.... in my thinking it should be (d/(v)) Yes I know that v=(x/t) however it was 5m/s or 3m/s so that means 5/1 or 3/1 but instead its getting flipped to be 1/5 or 1/3..... this stuff going on here is confusing the crud out of me... :(

Brian

 Wilder7bc Jan 19th 2010 09:21 AM

maybe hhmm...

Quote:
 Originally Posted by physicsquest (Post 11822) The average speed is defined as the total distance / total time. Let distance between A and B = d. Then total distance to and fro = 2d. The time one way = d/5 and the other way = d/3 . Total time therefore is d/5 + d/3 = 8d/15. Avg speed = 2d / (8d/15) = 3.75

Here is how I would have tried to do this and perhaps seeing how I would try to do it incorrectly you can figure out what I am thinking wrong.
(little hard to do this as I am at work as well, and I have to think pretty deep on things usually and can be easily distracted so please forgive any silly mistakes)

following what I think he said up above...

V =(2D/t) which I can change to:
Vt = 2D which I can change again to:
t =(2D/V)

I can then plug in what numbers I have?

t = (2D/(5m/1s + 3m/1s)

I get to this point and I dont know how to take it any further or not sure anyway.....

 Corum Jan 19th 2010 10:41 AM

Think along the lines that distance is (Xfinishing point - Xstarting point). allowing for V=0 or negative V.
Average speed = 2d / (8d/15), so:
= 2d*15/8d
=30d/8d= d(30)/d(8) = 1*(30/8)=3,75.

You can't plug in numbers in "t =(2D/V)" because d is unknown.

 YellowPeril Jan 19th 2010 11:40 AM

why?

Wilder7bc,

Look at this in terms of an average. As you know speed = distance / time.

If I had to tell you that a bus on its outbound route took 0.5 hour to do 1 km and on its return route took 0.75 hour to do 1 km then I am quite sure you will agree that the average speed is 2 km / (0.5 + 0.75) or 1.6 km per hour.

The formula for speed is distance / time so the speed out bound is 2 km / hour and the speed returning is 4/3 km per hour. If I were to give you only the speeds and the distances you would have to work out the time. Which would be an extra step to get the same answer.

In statistics there are different ways of calculating the mean and one is called the harmonic mean. It is simply n / (sum(1/vi)) where sum(1/vi) is the sum of the inverse of (in this case speed) if the speed is slower, the speed is traveled at for a longer time hence this type of weighting is often used when calculating average speeds.

I know it does not seem logical at first but it helps to view it as a weighted average by time.

 Wilder7bc Jan 19th 2010 11:45 AM

Quote:
 Originally Posted by Corum (Post 11846) Think along the lines that distance is (Xfinishing point - Xstarting point). allowing for V=0 or negative V. Average speed = 2d / (8d/15), so: = 2d*15/8d =30d/8d= d(30)/d(8) = 1*(30/8)=3,75. You can't plug in numbers in "t =(2D/V)" because d is unknown.
I understand everything that is being said till I see the magic....

I dont doubt its right whats being said but I cannot see the details of how it happens, so I am confused...

How do you get (8d/15) its actually:

(5m/1s + 3m/1s) What is the math steps that transform this into (8/15)

(5/1) + (3/1) = (8/1) NOT (8/15)

Somehow because of some math rule or law thats being left out I guess which is why I am calling it magic as has not been defined yet I keep seeing that people are taking

(5m/1s + 3m/1s

and turning it into

(1/5) + (1/3) then of course I understand the math from there they just are taking a common denominator and they end up with (8/15)

Its that Law or rule thats being left out that people are doing in their head or something that I am not understanding, and why I use the term magic.

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