Calculate Acceleration at t=38.60s: Simple Step-by-Step Guide

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To calculate the acceleration at t=38.60 s, the object’s position data at three time points is used. The change in position from 9.600 m to 15.804 m over the time interval from 37.40 s to 38.60 s allows for the determination of velocity. By applying the kinematic equation x = v0t + 0.5at^2, the initial velocity can be calculated. Substituting the known values into the equation will yield the acceleration. This method provides a straightforward approach to solving for acceleration using the given position data.
zhenyazh
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ok this is really easy. and i solved it once and now i can't get to the same solution.

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)
37.40 | 9.600
38.60 | 15.804
39.80 | 27.696
Calculate the magnitude of the acceleration at t=38.60 s.

so i tried with the x and v formuals, it gets me nowhere
also tried with finding inclines.

can u give me an outline of how to do it?

thanks
 
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you know that the object from time 37.4 till 38.6 (s) moved from 9.6 to 15.804(m) , can you get the velocity from these information then substitute in any of the kinematic equations( i.e x = vot + 0.5 at^2 ) ..
 

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