What is the acceleration of an object moving with a constant acceleration?

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Homework Help Overview

The problem involves an object moving in a straight line with constant acceleration, with position data provided at three different times. The goal is to calculate the magnitude of the acceleration at a specific time based on the given position measurements.

Discussion Character

  • Exploratory, Assumption checking

Approaches and Questions Raised

  • The original poster attempts to calculate acceleration using various equations but struggles with the application of time intervals and distance measurements. Some participants suggest clarifying the use of position data in the equations and propose setting up a system of equations based on the provided measurements.

Discussion Status

Participants are actively engaging with the problem, with some offering guidance on how to correctly apply the equations. The original poster expresses a realization about a mistake in their calculations, indicating progress in understanding the problem.

Contextual Notes

There is a mention of confusion regarding the time intervals used in calculations, specifically the difference between using 3 seconds versus 1.5 seconds for the acceleration calculation.

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Homework Statement



An object is moving in a straight line with a constant acceleration. Its position is measured at three different times, as shown in the table below.

Time (s) Position, (m)
37.10 7.600
38.60 14.425
40.10 27.100

Calculate the magnitude of the acceleration at t=38.60 s.


Homework Equations



a = (vf - vo) / t
d = volt + 1/2 at2
vf2 = vo2 + 2ad
(ether or really)

The Attempt at a Solution



I've tried everything, I've tried getting the distances between the 2 each time and using 1.5 seconds and getting velocity, I've tried using those to get the velocity, nothing is working. and this is something I should be able to do. Any help would be good, even just a point in the right direction.
 
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In the equation [tex]d=v_0t+\frac{at^2}{2}[/tex]
[tex]d[/tex] is a distance. But in the statement of the problem it says that the position of the object is measured...so you can't replace d with the data for the position.
But if you write the distance d as the difference between the position (say [tex]x[/tex]) measured at a time and the initial position (say [tex]x_0[/tex]) and replace in the equation, then you have:

[tex]x-x_0=v_0t+\frac{at^2}{2}[/tex]

Now all you have to do is replace the 3 sets of data in the equation above and you have a system of 3 equations with 3 unknowns (one of them being the acceleration).
 
Last edited:
Thanks but I just got it. Before I was using 3 seconds when I was suppose to be using 1.5s, what a silly mistake.

<I>d1</I> = 14.425-7.600 = 6.825 m / 1.5s(38.60 - 37.10) = <i>v1</I> = 4.55 m/s
<I>d2</I> = 27.100-14.425 = 12.675 m / 1.5s = <i>v2</I> = 8.45 m/s
<T> = 1.5s

V1-V2/T = 2.60 m/s^2
 
CoverlessTech said:
Thanks but I just got it. Before I was using 3 seconds when I was suppose to be using 1.5s, what a silly mistake.

<I>d1</I> = 14.425-7.600 = 6.825 m / 1.5s(38.60 - 37.10) = <i>v1</I> = 4.55 m/s
<I>d2</I> = 27.100-14.425 = 12.675 m / 1.5s = <i>v2</I> = 8.45 m/s
<T> = 1.5s

V1-V2/T = 2.60 m/s^2

That is also correct.
 

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