To determine the position of the object when the acceleration is zero, we will follow these steps:

1. Find the velocity $v(t)$ by taking the first derivative of the position function $P(t)$.

   $v(t) = \frac{d}{dt}(P(t)) = \frac{d}{dt}(t^3 - 9t^2 + 50t)$ $v(t) = 3t^2 - 18t + 50$

2. Find the acceleration $a(t)$ by taking the second derivative of the position function $P(t)$, or the first derivative of the velocity $v(t)$.

   $a(t) = \frac{d}{dt}(v(t)) = \frac{d}{dt}(3t^2 - 18t + 50)$ $a(t) = 6t - 18$

3. Set the acceleration to zero and solve for $t$.

   $6t - 18 = 0$ $6t = 18$ $t = 3$

4. Find the position at $t = 3$ by plugging $t = 3$ into the original position function $P(t)$.

   $P(3) = (3)^3 - 9(3)^2 + 50(3)$ $P(3) = 27 - 81 + 150$ $P(3) = 96$

So, the position of the object when the acceleration is zero is $96$ units.

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Would you like more details on any part of the solution? Here are five related questions you might find useful:

1. How do you interpret the velocity when the acceleration is zero?
2. What is the significance of finding where acceleration equals zero in physics?
3. How would the object's position change if the coefficients of the function were altered?
4. How can you find the object's maximum height using this function?
5. What is the importance of taking higher derivatives like acceleration in motion problems?

Tip: Remember that acceleration is the rate of change of velocity, and a zero acceleration often means that the velocity is constant at that instant.