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The equation of position vs. time for a moving object, in SI units, is as $x=-t^{2}+6 t-9 .$ Which of the following choices are correct?

(a) The object's acceleration is constant and its magnitude is $1 \mathrm{~m} / \mathrm{s}^{2}$.
(b) The object's velocity at the initial time $t=0$ is to the negative $x$-axis.
(c) The object's initial position is on the negative side of the $x$-axis.
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(a) This equation has a quadratic form so its acceleration is constant. Comparing this equation with standard constant acceleration kinematic equation, $x=\frac{1}{2} a t^{2}+v_{0} t+x_{0}$, we will find its magnitude as
$\frac{1}{2} a=-1 \Rightarrow a=-2 \mathrm{~m} / \mathrm{s}^{2}$
So this choice is incorrect.

(b) "Object's velocity at the initial time" means its initial velocity. Comparing the two equations below reveals that initial velocity is $v_{0}=+6 \mathrm{~m} / \mathrm{s}$.
$\begin{gathered} x=\frac{1}{2} a t^{2}+v_{0} t+x_{0} \\ x=-t^{2}+6 t-9 \end{gathered}$
The positive sign indicates that the initial velocity is toward the positive $x$-axis. So, this choice is also incorrect.

(c) By setting $t=0$ in the position-time equation, its initial position is obtained. So, $x_{0}=$ $-9 \mathrm{~m}$. Negative indicates that the object is on the negative side of the $x$-axis initially. Thus, this choice is correct.
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