Time to Acceleration: Convert Milliseconds?

  • Thread starter Thread starter doc.madani
  • Start date Start date
  • Tags Tags
    Acceleration Time
Join the discussion
Registration is free. Ask a follow-up in this thread, or start your own.
3 replies · 2K views
doc.madani
Messages
90
Reaction score
0
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
 
Physics news on Phys.org
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 ...