When is the particle speed up/slowing down

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The particle's position is defined by the equation s=2t^2-9t^2. The first derivative, representing velocity, is calculated as 6t^2-18t, and the second derivative, representing acceleration, is 12t-18. The particle speeds up when both velocity and acceleration share the same sign and slows down when they have opposite signs. To determine when this occurs, it's essential to find the values of t where the velocity and acceleration change signs, which happens at their respective zeros. The analysis of these derivatives will provide the necessary values of t for the particle's speed behavior.
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Homework Statement



The postion of a particle is given by s=2t^2-9t^2 where t is in seconds and s in meters

a) when is the particle speeding up/slowing down

Homework Equations





The Attempt at a Solution



I have calculated the
1st derivative (velocity) as 6t^2-18t
2nd derivative (acceleration) as 12t-18

and I know that the particle is speeding up when the velocity and the acceleration have the same sign and slowing down when they have opposite signs. However I am not sure how to express this to answer the question.

Thanks
 
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Find the values of t for which the velocity and for which the acceleration is positive/negative. Hint: They can only change sign at the zeros.
 

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