Extended Experimental Investigation.

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Homework Help Overview

The discussion revolves around an Extended Experimental Investigation in physics, focusing on the relationship between load, displacement, velocity, and acceleration of a trolley using ticker tape. The original poster, Aaron, expresses uncertainty about measuring displacement after forgetting to take that measurement during the experiment.

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • Participants inquire about the measurements Aaron has taken and the methodology used to derive average acceleration values from the ticker tape data. There is a focus on clarifying how the distances between marks were measured and how these relate to the overall experimental setup.

Discussion Status

The discussion is ongoing, with participants providing clarifications and exploring the implications of the data Aaron has collected. Some guidance has been offered regarding the calculation of average acceleration and the potential construction of graphs to analyze displacement and velocity.

Contextual Notes

Aaron mentions missing information due to a hospital stay, which may affect his understanding of the experiment. There is also uncertainty regarding the angle of the ramp used in the experiment, which could influence the results.

  • #31
No. I don't really have any clue what your talking about there, sorry.
 
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  • #32
On order to draw the graph you need to generate the coordinates for the datapoints using the data. For the no load graph (0 gram) the displacement was 2.0 cm after 0.1 seconds (time for 5 intervals). Another 0.1 seconds later is was 5.7 centimeters further away, so the coordinates is 0.2 seconds and 5.7 + 2.0 cm = 7.7 cm
 
  • #33
So..
0.1 seconds = 2cm
0.2 seconds = 7.7cm (5.7cm + 2.0cm)
0.3 seconds = 17.5cm (9.8cm + 5.7cm + 2.0cm)
0.4 seconds = 31.5cm (14.0cm + 9.8cm + 5.7cm + 2.0cm)
0.5 seconds = 48.9cm (17.4cm + 14.0cm + 9.8cm + 5.7cm + 2.0cm)
0.6 seconds = 69.9cm (21.0cm + 17.4cm + 14.0cm + 9.8cm + 5.7cm + 2.0cm)
0.7 seconds = 89.3cm (19.4cm + 21.0cm + 17.4cm + 14.0cm + 9.8cm + 5.7cm + 2.0cm)

Does that look right?
 
  • #34
Correct. To generate the average velocity graph you would determine the gradient of this graph by using pairs of successive data points on the displacement graph. This means that the delta y's will be the additional distance added to get the displacement and the delta x will be 0.1 second each time. We say that the average velocity was in the middle of each of these intervals, so that would be at 0.15 seconds, 0.25 seconds. This means that the first two point will be

(0.15\ s,\ 57\ cm/s)

and

(0.25\ s,\ 98\ cm/s)
 
  • #35
I don't get why we start at 0.15 s and not at like 0.05 s. Plus, how did you get 57cm/s?

(Try to dumb things down a lot hehe) I don't really know what half of this stuff is such as delta y's, gradient and delta x.
 
  • #36
The first average velocity is obtained by using the data points

0.1\ s,\ 2.0\ cm

and

0.2\ s,\ 7.7\ cm

giving

\bar{v} = \frac{7.7 - 2.0}{0.2 - 0.1} = 57\ cm/s

This will be the average velocity in the middle of 0.1 and 0.2 seconds, that is at 0.15 seconds.
 
  • #37
I'm getting confused again. I don't even remember/know why we're trying to work out 0.15 seconds. =S

Anyway, I've got to go to bed so i'll be back tomorrow morning or tomorrow afternoon.

Thanks for helping. =)
 
  • #38
The displacements of the trolley occurred at 0.1 and 0.2 seconds during the recording. This means that the calculated average velocity will be in the middle of this period, that is it will be the average velocity of the trolley at 0.15 seconds.

It is only 15:45 here in South Africa.
 
  • #39
Ohh. I'm with you now. Yeah that would make sense seeing as I need to add the average velocity also. It was like 11pm when I went to bed. Big time different.
 
  • #40
Does anyone know what the formula is for working out the velocity or displacement?
 

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