Ticker-Tape Cart, moving down ramp Experient, determine acceleration w/o graphs

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SUMMARY

The discussion focuses on determining the acceleration of a cart rolling down a ramp using a ticker tape timer, specifically without relying on graphing methods. The participant has collected position-time data and established that the cart experiences uniform acceleration. They attempted to calculate acceleration using the equation a = (2∆d)/(∆t²) but obtained an incorrect value of 40 m/s², while graphing indicated 24 m/s². The conversation emphasizes the importance of accurate measurements and averaging multiple data points to achieve a reliable acceleration value.

PREREQUISITES
  • Understanding of kinematic equations, specifically ∆v = ∆d/∆t and a = (vf - vi)/∆t
  • Familiarity with ticker tape experiments and data collection methods
  • Knowledge of uniform acceleration concepts
  • Basic skills in averaging data for improved accuracy
NEXT STEPS
  • Research methods for averaging measurements in physics experiments
  • Learn about error analysis and its impact on experimental results
  • Explore advanced kinematic equations for varying acceleration scenarios
  • Investigate the use of digital timing tools for improved accuracy in experiments
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Students in physics, educators conducting experiments on motion, and anyone interested in improving accuracy in kinematic data collection and analysis.

hoho1
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I have just done this experiment where i have a cart roll down a ramp and a ticker tape timer does its ticks on a tape atached to the cart.
already i can make a postion (or displacement from a dot of reference)- time table. for 1.0 second going down the ramp
i all ready have finished the questions and already know the carts moving with uniform acceleration. i have created position-time, velocity-time and acceleration-time graphs as well.

the question is "graphing is suggested for determining the acceleration. however, it is possible to apply the defining equation for average acceleration to determine the acceleration of the moving object... describe how you would do this, including the assumption(s) that you must make to solve the problem. (hint: how would you obtain a fairly accurate vale of final velocity?)"


Homework Equations


∆v =∆d/∆t a=∆v/∆d a=(vf-vi)/∆t


The Attempt at a Solution


because i can only use infomation not obtained through graphing i have that position-time table [at 0.0m, 0.0s; 1.85m, 0.2s; 5.0m,0.4s; 9.10m,0.6s; 14.1m, 0.8s; 20.0m,1.0s] and knowlegde its a constant acceleration problem. i chose a 0.0m marker at the first clear dot on the ticker tape, so where i choose it doesn't affect the results.

my try was intial velocity was 0m/s, change in time 1.0s, displacement 20.0cm

∆d=(vi ) ∆t+(a ∆t2)/2 which because of the above ∆d=(a(∆t)2)/2
then a=(2∆d)/(∆t2) but this results in the value of 40m/s2? this can't be right because my value that i got from the graphs gave me 24m/s2.

help needed for another way to solve this or find final velocity

oh yeah if no one knows ∆ is change in just in case, d is displacement, v is velocity, a is acceleration, t is time in s
 
Last edited:
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Welcome to PF, hoho.
The problem here is that there is error in the measurements - rounding the distance to the nearest 1 mm makes quite a difference when you take the difference between two such values. You need to average over many dots. The usual way to do this is with a table like this:
tttable.jpg

These are not your numbers - just an example from another tickertape experiment. Notice that you get wildly different numbers in the acceleration column, but averaging that last column will provide a reasonable value for the acceleration. I see I made a mistake on the last number - should be -300, not 300.
 

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