Position/acceleration derivatives problem

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SUMMARY

The object described by the position function s(t) = 4t³ - 3t² + 5 travels along a linear path. The acceleration function, derived as a(t) = 24t - 6, indicates that the acceleration is zero at t = 1/4 seconds. By evaluating the position function at this time, the distance traveled from the initial position of 5 meters is calculated as d = s(1/4) - s(0) = -1/8 meters. This confirms the calculations are correct.

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Homework Statement



An object is traveling along a linear path according to the equation s(t) = 4t^3 - 3t^2 + 5 where t is measured in seconds and s(t) measured in meters. How far has the object traveled when its acceleration is zero?

Homework Equations





The Attempt at a Solution



The acceleration time function is (or second derivative of position) is, a(t)= 24t-6.
Then since a(t)=0, t=1/4 s.
And know we take the difference between s(1/4) and s(0), since the object initially starts 5m to the right of the origin at t=0s.
So, d=s(1/4)-s(0)
=-1/8m
Is this correct?
 
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