Time to Acceleration: Convert Milliseconds?

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SUMMARY

The discussion centers on the conversion of time from milliseconds to seconds when calculating acceleration using the formula \( a = \frac{2x}{t^2} \). It is established that if time is in milliseconds, it must be converted to seconds by dividing by 1000 before substituting into the formula for acceleration in meters per second squared (m/s²). Additionally, it is clarified that a distance-time graph for accelerated motion should be represented as a parabola rather than a line of best fit, particularly when acceleration is not constant.

PREREQUISITES
  • Understanding of basic physics concepts, specifically acceleration
  • Familiarity with the formula for acceleration \( a = \frac{2x}{t^2} \)
  • Knowledge of unit conversion, particularly between milliseconds and seconds
  • Ability to interpret distance-time graphs and the significance of parabolic shapes
NEXT STEPS
  • Study the derivation and application of the acceleration formula \( a = \frac{2x}{t^2} \)
  • Learn about unit conversions in physics, focusing on time measurements
  • Explore the characteristics of distance-time graphs for different types of motion
  • Investigate the mathematical integration of acceleration to derive position functions
USEFUL FOR

Students of physics, educators teaching kinematics, and anyone involved in experimental physics or motion analysis will benefit from this discussion.

doc.madani
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hello just a quick question, in the experiemental formula for acceleration (2xr/t^2) if the time was in milli seconds e.g 0.00.67 seconds, before u substitute the value in the formula would u need to convert that time to 1 second, by timing everything by 1000 then diving the values by 6.7 to find the acceleration for that 1 second? (m/s ^ -2)


thankyou
 
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and another question in a distance vs tme graph would i need to use a line of best fit or can i use a parabola, because the speed is obviously accelerating
 
doc.madani said:
hello just a quick question, in the experiemental formula for acceleration (2xr/t^2) if the time was in milli seconds e.g 0.00.67 seconds, before u substitute the value in the formula would u need to convert that time to 1 second, by timing everything by 1000 then diving the values by 6.7 to find the acceleration for that 1 second? (m/s ^ -2)
Where did you get that formula from? Because it doesn't seem to be correct.

But anyway, if you had a correct formula for acceleration, and you had time in milliseconds, and you had acceleration in meters per second squared, then yes you would have to use a conversion factor from milliseconds to seconds. But if you had time in milliseconds and acceleration in meters per millisecond squared, you would not have to use a conversion factor.

In a distance-time graph, accelerated motion looks like a parabola, not a line. So you could not use a best-fit line.
 
doc.madani said:
and another question in a distance vs tme graph would i need to use a line of best fit or can i use a parabola, because the speed is obviously accelerating

I don't understand what you mean by line of best fit. You can use a parabola only if the acceleration is constant. Otherwise, you will need to find a mathematical expression for x(t) using the known form of the acceleration and integrating twice. Here it seems that the acceleration is not constant so ...
 

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