Extended Experimental Investigation.

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SUMMARY

The forum discussion centers on an Extended Experimental Investigation involving physics concepts such as displacement, velocity, and acceleration using a ticker timer. The user, Aaron, seeks assistance in calculating displacement after forgetting to measure it during the experiment. He provides detailed measurements from multiple ticker tapes and discusses how to derive average acceleration and displacement values based on time intervals and mass loads applied to a trolley. Key calculations involve using the formula for average velocity and understanding the relationship between displacement and time.

PREREQUISITES
  • Understanding of basic physics concepts: displacement, velocity, and acceleration
  • Familiarity with ticker timers and their operation
  • Ability to interpret and analyze experimental data
  • Knowledge of graphing techniques for plotting displacement versus time
NEXT STEPS
  • Learn how to calculate average velocity using displacement data
  • Explore graphing techniques for visualizing displacement and velocity
  • Study the principles of kinematics related to motion and acceleration
  • Investigate the effects of mass on acceleration in physics experiments
USEFUL FOR

Students in physics courses, particularly those conducting experiments involving motion, as well as educators seeking to guide students through data analysis and interpretation in experimental investigations.

  • #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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