Curse

Toxage · Sep 23, 2008, 07:48 PM // 19:48

Graph :

http://img354.imageshack.us/my.php?image=physicslp1.png

or

http://www.turboimagehost.com/p/766378/physics.JPG.html

Some people had trouble with imageshack.

Questions:

Determine its average velocity between the following times.

1. t = 25.0 s and t = 30.0s

2. t = 35.0 s and t = 50.0 s

3. The attempt at a solution

average velocity = overall displacement / overall time

1. 6m/5 sec = 1. m/s

2. 10 m / 15 s = 2/3 ms

My teacher uses on online homework program called Webassign, and both of these showed up as wrong. I don't know why. If anyone could help me that would be awesome.

Jumping Is Uselss · Sep 23, 2008, 07:56 PM // 19:56

I use webassign too, but for calculus.

Do you have the distances for us to use?

Toxage · Sep 23, 2008, 08:11 PM // 20:11

Quote:

Originally Posted by Jumping Is Uselss

I use webassign too, but for calculus.

Do you have the distances for us to use?

I provided the graph for the distances. I'm confused...

Jumping Is Uselss · Sep 23, 2008, 08:24 PM // 20:24

Find the distance it travels in 25sec and in 30 secs. I had to estimate it, but it came out 25 sec = 8m and 30sec = 16m.

V= change in distance/change in time. (30-25)/(16-8) = .625

Could be wrong, long time since ive done physics. And not really at the top my list for my major.

FlamingMetroid · Sep 23, 2008, 08:33 PM // 20:33

dont help Toxage >_>

Quote:

Originally Posted by Jumping Is Useless

Find the distance it travels in 25sec and in 30 secs. I had to estimate it, but it came out 25 sec = 8m and 30sec = 16m.

V= change in distance/change in time. (30-25)/(16-8) = .625

Could be wrong, long time since ive done physics. And not really at the top my list for my major.

I'll go ahead and second that though

RedStar · Sep 24, 2008, 03:30 AM // 03:30

Never mind, got the wrong calcul xD.

You are normally right

Lord Sojar · Sep 24, 2008, 03:44 AM // 03:44

The idea is that you are calculating position as related to time, which is what defines velocity. To understand the problem, you must understand velocity itself. Velocity is the time from a point to another point based on time and spacial movement, quite different from speed.

You should really just give him the equations to work with and let him do it on his own. I generated this formula using mathematica for you Toxage. Using what I just told you, I am sure you can figure out what each variable represents and how to use them. By using these and plugging in the values from each interval, you can derive the answers.

RedStar · Sep 24, 2008, 03:51 AM // 03:51

Great another formula to calculate the same thing T_T

Rahja, what's the difference between what you have posted, this :
Vavg=d(i)-d(i-1)/t

and this :

Vavg=d(i+1)-d(i-1)/2*t

?

Lord Sojar · Sep 24, 2008, 03:57 AM // 03:57

I am being particular... leave me alone!

You used a divisor in your formula, which are generally not accepted on exams because they are not standard form (university level at least, you can get away with murder in HS physics)

But yes, the formulas work the same way. Most of the time, these issues come up as a result of people not really understanding what time is in respect to spacial movement and dimension. You seem to grasp it. (Also, it helps if you use mathematica or the like to image formulas to ease confusion with students, as seeing a formula in proper form does wonders for comprehension.)

EDIT: By divisor, I mean numeric non standard divisor (aka that pesky 2)

RedStar · Sep 24, 2008, 04:11 AM // 04:11

I see (I think xD).

Well I put the formula like our teacher gave it to us (except for the / which is supposed to be _____ ). Maybe it's because she didn't want to make it seem too complicated for us lol.

Turbobusa · Sep 24, 2008, 05:54 AM // 05:54

Looks like a sin function between 35 and 50, rounding it to a constant deceleration would be a bit much imo.
I'd suggest derivating (proper english?) that for that period.

The 25-30s period can be rounded to a straight line and thus you can safely use d(x)/d(y) there.

Btw from post #8 I'm not sure if you understand what a derivative function is, I'd try to understand that first.

RedStar · Sep 24, 2008, 08:23 AM // 08:23

Sorry, I can speak fluent english, but I have never learned the math and physics terms...

From what information he gave us and what results he wanted, I thought that his exercise was to calculate the Average velocity between 2 points, because a few weeks ago we had an exercise really similar to this one.
Unless "Vitesse Moyenne" and "Average Velocity" are not the same terms.

**Marty Silverblade** · Sep 24, 2008, 09:07 AM // 09:07

Quote:

Originally Posted by Turbobusa

I'd suggest derivating (proper english?) that for that period.

Deriving

.

Turbobusa · Sep 24, 2008, 11:27 AM // 11:27

Quote:

Originally Posted by RedStar

From what information he gave us and what results he wanted, I thought that his exercise was to calculate the Average velocity between 2 points, because a few weeks ago we had an exercise really similar to this one.

Oui c'est bien ce qu'il veut, mais la position est l'intégrale de la vitesse

Donc si tu dérive la position (tu calcules la différence de position entre 2 instants donnés), tu obtiens la vitesse, moyenne puisque la vitesse instantanée est plutot calculée avec 2 points a des instants très rapprochés.
Si j'ai parlé de fonction sinus c'est parce qu'il ne devrait pas appliquer la meme formule pour les 2 périodes puisque dans la seconde période la position ne varie pas de facon linéaire.

Tes formules sont bonne mais il faut essayer de comprendre ce qu'elles représentent, ca aide beaucoup a la compréhension globale.

--

Indeed this is what I understood he wanted, and knowing that position is the integral of velocity, if you derive the position diagram you get the speed variation. As you derive from two points separated by more than 1 unity of time, it is an average.
If I mentionned the sin function it is because he shouldn't apply the same formula to both periods seing that in the second period the position variation is not linear.

Your formulas are right, but you need to understand them to get a better overall comprehension.

--

Note:
dx/dt = Velocity
dv/dt = dx/dt² = Acceleration

Quote:

Originally Posted by Marty Silverblade

Deriving

.

Thank you

Divinitys Creature · Sep 24, 2008, 09:30 PM // 21:30

Quote:

Originally Posted by Toxage

Graph :

http://img354.imageshack.us/my.php?image=physicslp1.png

or

http://www.turboimagehost.com/p/766378/physics.JPG.html

Some people had trouble with imageshack.

Questions:

Determine its average velocity between the following times.

1. t = 25.0 s and t = 30.0s

2. t = 35.0 s and t = 50.0 s

3. The attempt at a solution

average velocity = overall displacement / overall time

1. 6m/5 sec = 1. m/s

2. 10 m / 15 s = 2/3 ms

My teacher uses on online homework program called Webassign, and both of these showed up as wrong. I don't know why. If anyone could help me that would be awesome.

You basically read off numbers off the graph poorly your teacher would have assumed. You should give an estimate of your error, and this is dominated by the poor calibration of the graph scales. If you did give an estimate of your error then you may well have got the right answer.

Since there are no error bars on the data points shown in the graph I think this is the sort of question where you just need a better eye.

For the first question the distance moved is more like 8 m in 5 s and so the average velocity is 8/5= +1.6 m/s.

For the second question the distance moved is -9 m in 15 s and so the average velocity is -0.6 m/s.

Notice how you missed the minus sign in your answer for the second part! That means you're saying the motion is in the opposite direction to which it actually was LOL.

illidan009 · Sep 25, 2008, 12:02 AM // 00:02

Quote:

Originally Posted by Turbobusa

Oui c'est bien ce qu'il veut, mais la position est l'intégrale de la vitesse

Donc si tu dérive la position (tu calcules la différence de position entre 2 instants donnés), tu obtiens la vitesse, moyenne puisque la vitesse instantanée est plutot calculée avec 2 points a des instants très rapprochés.
Si j'ai parlé de fonction sinus c'est parce qu'il ne devrait pas appliquer la meme formule pour les 2 périodes puisque dans la seconde période la position ne varie pas de facon linéaire.

Tes formules sont bonne mais il faut essayer de comprendre ce qu'elles représentent, ca aide beaucoup a la compréhension globale.

--

Indeed this is what I understood he wanted, and knowing that position is the integral of velocity, if you derive the position diagram you get the speed variation. As you derive from two points separated by more than 1 unity of time, it is an average.
If I mentionned the sin function it is because he shouldn't apply the same formula to both periods seing that in the second period the position variation is not linear.

Your formulas are right, but you need to understand them to get a better overall comprehension.

--

Note:
dx/dt = Velocity
dv/dt = dx/dt² = Acceleration

Thank you

Sweeet I can read your French! 4 years NOT gone to waste.
Yeah seems like some ppl beat me to the phyziks problem.

RedStar · Sep 25, 2008, 03:10 AM // 03:10

Thank you for using french ^^.
Well I still have 2 more years to really understand all of that

.