Newbie needs basic help with magnitude slope

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    Magnitude Slope
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SUMMARY

The discussion focuses on calculating the magnitude of the maximum slope from an experiment measuring instantaneous speed while driving a car. The user correctly applies the formula for slope, using rise over run, resulting in an acceleration of 1.38 m/s². This value is then compared to the acceleration due to gravity, which is 9.8 m/s², concluding that the calculated acceleration is less than that of gravity. The user seeks clarification on unit conversions and the representation of acceleration in their findings.

PREREQUISITES
  • Understanding of basic physics concepts, specifically acceleration.
  • Familiarity with unit conversions, particularly from km/h to m/s.
  • Knowledge of graphing techniques, including plotting time vs. speed.
  • Basic mathematical skills for calculating slope using rise over run.
NEXT STEPS
  • Learn about unit conversions between km/h and m/s for accurate calculations.
  • Study the concept of acceleration and its representation in physics.
  • Explore how to express acceleration in terms of gravitational force (g's).
  • Review graphing techniques for better visualization of speed vs. time data.
USEFUL FOR

Students in introductory physics courses, educators teaching basic kinematics, and anyone interested in understanding the principles of acceleration and unit conversions.

urbano
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Homework Statement



I have done an experiment where every 10 seconds I had to say my instantaneous speed while drivng a car and my co-passenger would record the time (we carried this out over 20 minutes). I then plotted this out on a graph with time (s) on the X axis and km/h on the Y axis.

I now need to calculate the magnitude of the maximum slope and then make a comparison statement about my result vs. the rate of acceleration due to gravity. ...and to be honest this is all new to me and I'm not sure if I'm doing this right.

Homework Equations



This is what I went with

y1= 0km/h, y2=50km/h, x1= 0s, x2= 10s. I then used rise over run (y2-y1)/(x2-x1)
but I'm not sure if this is the right formula or what you'd call the resulting units...km/h or maybe convert to m/s or m/s^2 ? and I'm not sure what it represents ? Acceleration (change of velocity/ time)??

The Attempt at a Solution



50-0/10-0 = 5 (somethings) ...then I thought maybe 50km/h should be put into m/s which I came up with 13.88 m/s...meaning (13.88 m/s) /( 10s) = 1.38 m/s or maybe it should be 1.38 ms^2.

so when i then compare this to the acceleration rate due to gravity (9.8m/s^2). I will make a statement my acceleration rate is less than that of gravity.

Any help anyone is willing to offer wold be most appreciated.

TIA
 
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urbano said:
y1= 0km/h, y2=50km/h, x1= 0s, x2= 10s. I then used rise over run (y2-y1)/(x2-x1)
but I'm not sure if this is the right formula or what you'd call the resulting units...km/h or maybe convert to m/s or m/s^2 ? and I'm not sure what it represents ? Acceleration (change of velocity/ time)??
Yes, acceleration. It's useful to keep track of units by keeping them in the equations:
50km/h / 10s = 5 km/h/s = 5000 m/h/s = 5000/3600 m/s2
so when i then compare this to the acceleration rate due to gravity (9.8m/s^2). I will make a statement my acceleration rate is less than that of gravity.
My guess is that you are expected to say how many g's it is.
 
cool, thanks for the help. We haven't touched on g's yet...(I don't think) . We are at are 101 stage for those with little or no background,so I imagine the g's may be not be too far off.
 

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